Business Calculus

Continuously compounding interest formula with examples

Continuously compounding interest is similar to regular compound interest however, interest is not compounded monthly or quarterly but instead, continuously. The continuously compounding interest formula can be used to find the future value of an investment at a given rate or the amount of time it takes to reach a future value given a desired amount. This is commonly taught in college algebra courses or sometimes calculus courses. Below we will look at the formula and some examples of using it.

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Examples of finding the future value with the continuously compounding interest formula

We will start with using the formula to calculate the future value of an investment. Then we will look at how to find the time it takes to reach a given future value.

Remember that $e$ is a constant. This is located on your calculator and you should use that instead of an approximation in order to get the most accurate answer.

Example

An investment of $12,000 is invested at a rate of 3.5% compounded continuously. What is its value after 6 years?

Solution

Determine what values are given and what values you need to find. Then, use the formula.

Investment of $12,000 – this is the principle: $P=12000$
Rate 3.5% – remember to write this as a decimal (divide by 100): $r = 0.035$
Value after 6 years: $t = 6$

You are finding the value 6 years in the future, so you are finding $A$, the future value.

$\begin{align}A &= Pe^{rt}\\ &= 12000e^{0.035\times6}\\ &= \bbox[border: 1px solid black; padding: 2px]{14,804.14}\end{align}$

Answer: The value will be $14,804.14 after 6 years.

On your calculator, if you are using a TI83 or TI 84, you should have the following.

The $e$ is located just above $LN$.

Example of finding the time to reach a certain value using the continuously compounding interest formula

In some problems, you may be given a goal value such as $10,000 and asked how long it will take to reach that value given an initial investment. Here, you will use the fact that $\ln(x)$ is the inverse of $e^x$.

Example

An investment of $5,500 is made at a continuously compounding rate of 2%. How many years will it take to reach a value of $15,000?

Solution

Again, review the values you know and what you are trying to find.

Investment of $5,500 so $P = 5500$
Rate 2% so $r = 0.02$
Reach a value of $15,000 – this is the future value so $A = 15000$

We are asked the time, $t$. Now use the formula. Recall the statement above that $\ln(x)$ is the inverse of $e^x$.

$\begin{align}A &= Pe^{rt}\\ 15000 &= 5500e^{0.02\times t}\\ \frac{15000}{5500} &= e^{0.02\times t}\\ \frac{30}{11} &= e^{0.02\times t}\\ \ln\left(\frac{30}{11}\right) &= \ln\left(e^{0.02\times t}\right)\\ \ln\left(\frac{30}{11}\right) &= 0.02t\\ \frac{\ln\left(\frac{30}{11}\right)}{0.02} &= t \\ \bbox[border: 1px solid black; padding: 2px]{50.17} &= t \end{align}$

Answer: About 50 years

Important! Wait as long as possible to round. Rounding early will make your answer off by even a couple of years. Do your work in the calculator. This applies to finding the future value as well.

Continue your study of interest

Now that you have studied the continuously compounding interest formula, you can study other types of interest. Such as:

The derivative of lnx and examples

Part of calculus is memorizing the basic derivative rules like the product rule, the power rule, or the chain rule. One of the rules you will see come up often is the rule for the derivative of lnx. In the following lesson, we will look at some examples of how to apply this rule to finding different types of derivatives. We will also see how using the laws of logarithms can help make taking these kinds of derivatives even easier.

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Examples of finding the derivative of lnx

image showing the rule that the derivative of lnx is 1/x

Remember that when taking the derivative, you can break the derivative up over addition/subtraction, and you can take out constants. This allows us to find the following.

Example

$\left(3\ln(x)\right)^{\prime} = 3\left(\dfrac{1}{x}\right) = \dfrac{3}{x}$

Example

$\left(\dfrac{\ln(x)}{5}\right)^{\prime} = \dfrac{1}{5}\left(\ln(x)\right)^{\prime} = \left(\dfrac{1}{5}\right)\left(\dfrac{1}{x}\right) = \dfrac{1}{5x}$

Example

$\left(2x^2 – \ln(x)\right)^{\prime} = 4x – \dfrac{1}{x}$

These show you the more straightforward types of derivatives you can find using this rule. But, if we combine this with the laws of logarithms we can do even more.

Using the laws of logarithms to help

For some derivatives involving ln(x), you will find that the laws of logarithms are helpful. In terms of ln(x), these state:

image showing the laws of logarithms for natural log lnx

Using these, you can expand an expression before trying to find the derivative, as you can see in the next few examples. Here, we will do into a little more detail than with the examples above.

Example

Find the derivative of the function:
$y = \ln(x^2)$

Solution

Before applying any calculus rules, first expand the expression using the laws of logarithms. Here, we can use rule (1). This step is all algebra; no calculus is done until after we expand the expression.

$y = \ln(x^2) = 2\ln(x)$

Now, take the derivative. This is the calculus step.

$\begin{align} y^{\prime} &= \left(2\ln(x)\right)^{\prime}\\ &= 2\left(\ln(x)\right)^{\prime}\\ &= 2\left(\dfrac{1}{x}\right)\\ &= \boxed{\dfrac{2}{x}}\end{align}$

In the example above, only one rule was needed to fully expand the expression. The next example shows you how to apply more than one rule.

Example

Find the derivative of the function.
$y = \ln(5x^4)$

Before taking the derivative, we will expand this expression. Since the exponent is only on the x, we will need to first break this up as a product, using rule (2) above. Then, we can apply rule (1).

$y = \ln(5x^4) = \ln(5) + \ln(x^4) = \ln(5) + 4\ln(x)$

Now take the derivative of the expanded form of the function, and then simplify.

$\begin{align} y^{\prime} &= \left(\ln(5) + 4\ln(x)\right)^{\prime}\\ &= \left(\ln(5)\right)^{\prime} + 4\left(\ln(x)\right)^{\prime}\\ &= 4\left(\dfrac{1}{x}\right)\\ &= \boxed{\dfrac{4}{x}}\end{align}$

You may be wondering what happened to $\ln(5)$. Remember – this is a constant. If you check your calculator, you will find that $\ln(5) \approx 1.61$. The derivative of a constant is zero.

None of these examples have used rule (3), so let’s look at one more example to see how that might be applied.

Example

Find the derivative of the function.
$y = \ln\left(\dfrac{6}{x^2}\right)$

Here we have a fraction, which we can expand with rule (3), and then a power, which we can expand with rule (1). Remember that this is just algebra – no calculus is involved just yet.

$y = \ln\left(\dfrac{6}{x^2}\right) = \ln(6) – \ln(x^2) = \ln(6) – 2\ln(x)$

Now that we have $\ln(x)$ by itself, we can apply the derivative rule for the natural log.

$\begin{align}y^{\prime} &= \left(\ln(6) – 2\ln(x)\right)^{\prime}\\ &= \left(\ln(6)\right)^{\prime} – 2\left(\ln(x)\right)^{\prime}\\ &= -2\left(\dfrac{1}{x}\right)\\ &= \boxed{-\dfrac{2}{x}}\end{align}$

As in the previous example, $\ln(6)$ is a constant, so its derivative is zero.

Combining with other rules

Each of the derivatives above could also have been found using the chain rule. As you study calculus, you will find that many problems have multiple possible approaches. However, there are some cases where you have no choice. For example, consider the following function.

$y = \ln(3x^2 + 5)$

Since this is not simply $\ln(x)$, we cannot apply the basic rule for the derivative of the natural log. Also, since there is no rule about breaking up a logarithm over addition (you can’t just break this into two parts), we can’t expand the expression like we did above. Instead, here, you MUST use the chain rule. Let’s see how that would work.

Example

Find the derivative of the function.
$y = \ln(3x^2 + 5)$

Apply the chain rule.

$\begin{align}y^{\prime} &= \dfrac{1}{3x^2 + 5}\left(3x^2 + 5\right)^{\prime}\\ y^{\prime} &= \dfrac{1}{3x^2 + 5}\left(6x\right)\end{align}$

Since this cannot be simplified, we have our final answer.

$\begin{align}y^{\prime} &= \dfrac{1}{3x^2 + 5}\left(6x\right)\\ &= \boxed{\dfrac{6x}{3x^2+5}}\end{align}$

In some problems, you will find that there is a bit of algebra in the last step, with common factors cancelling. Be sure to always check for this.

Summary

Remember the following points when finding the derivative of ln(x):

The derivative of $\ln(x)$ is $\dfrac{1}{x}$.
In certain situations, you can apply the laws of logarithms to the function first, and then take the derivative.
Values like $\ln(5)$ and $\ln(2)$ are constants; their derivatives are zero.
$\ln(x + y)$ DOES NOT EQUAL $\ln(x) + \ln(y)$; for a function with addition inside the natural log, you need the chain rule.
$\ln(x – y)$ DOES NOT EQUAL $\ln(x) – \ln(y)$; for a function with subtraction inside the natural log, you need the chain rule.

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Continue your study of derivatives

The quotient rule

There are many so-called “shortcut” rules for finding the derivative of a function. The quotient rule, is a rule used to find the derivative of a function that can be written as the quotient of two functions. More simply, you can think of the quotient rule as applying to functions that are written out as fractions, where the numerator and the denominator are both themselves functions.

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Table of contents:

The rule
Remembering the quotient rule
Examples of using the quotient rule
Hints: saving work by simplifying first

The formula for the quotient rule

For functions f and g, and using primes for the derivatives, the formula is:

The quotient rule. For f over g, it is numerator f prime g minus f g prime and denominator g squared.

Remembering the quotient rule

You can certainly just memorize the quotient rule and be set for finding derivatives, but you may find it easier to remember the pattern. This is shown below.

Trick to remember the quotient rule. Think of the numerator as the top and the denominator as the bottom. Write the product TB out twice and subtract. Then, prime on the first and prime on the second (this gives you T prime B minus T B prime. Then, divide by B squared.

Examples

Naturally, the best way to understand how to use the quotient rule is to look at some examples. Notice that in each example below, the calculus step is much quicker than the algebra that follows. This is true for most questions where you apply the quotient rule. You will often need to simplify quite a bit to get the final answer.

Example

Find the derivative of the function:
$f(x) = \dfrac{x-1}{x+2}$

Solution

This is a fraction involving two functions, and so we first apply the quotient rule.

$f^{\prime}(x) = \dfrac{(x-1)^{\prime}(x+2)-(x-1)(x+2)^{\prime}}{(x+2)^2}$

Find the derivative for each prime.

$f^{\prime}(x) = \dfrac{(1)(x+2)-(x-1)(1)}{(x+2)^2}$

Simplify, if possible.

$\begin{align}f^{\prime}(x) &= \dfrac{(x+2)-(x-1)}{(x+2)^2}\\ &= \dfrac{x+2-x+1}{(x+2)^2}\\ &= \boxed{\dfrac{3}{(x+2)^2}}\end{align}$

When applying this rule, it may be that you work with more complicated functions than you just saw. In the next example, you will need to remember that:

$(\ln x)^{\prime} = \dfrac{1}{x}$

To review this rule, see: The derivative of the natural log

Example

Find the derivative of the function:
$y = \dfrac{\ln x}{2x^2}$

Solution

As above, this is a fraction involving two functions, so:
Apply the quotient rule.

$y^{\prime} = \dfrac{(\ln x)^{\prime}(2x^2) – (\ln x)(2x^2)^{\prime}}{(2x^2)^2}$

Find the derivative for each prime.

$y^{\prime} = \dfrac{(\dfrac{1}{x})(2x^2) – (\ln x)(4x)}{(2x^2)^2}$

Simplify, if possible.

$\begin{align}y^{\prime} &= \dfrac{2x – 4x\ln x}{4x^4}\\ &= \dfrac{(2x)(1 – 2\ln x)}{4x^4}\\ &= \boxed{\dfrac{1 – 2\ln x}{2x^3}}\end{align}$

Hint: Save work by simplifying first

Once you have the hang of working with this rule, you may be tempted to apply it to any function written as a fraction, without thinking about possible simplification first. This could make you do much more work than you need to! Consider the following example.

Example

Find the derivative of the function:
$g(x) = \dfrac{1-x^2}{5x^2}$

Given the form of this function, you could certainly apply the quotient rule to find the derivative. However, we can apply a little algebra first. Since the denominator is a single value, we can write:

$g(x) = \dfrac{1-x^2}{5x^2} = \dfrac{1}{5x^2} – \dfrac{x^2}{5x^2} = \dfrac{1}{5x^2} – \dfrac{1}{5}$

Now, using the definition of a negative exponent:

$g(x) = \dfrac{1}{5x^2} – \dfrac{1}{5} = \dfrac{1}{5}x^{-2} – \dfrac{1}{5}$

Now we can apply the power rule instead of the quotient rule:

$\begin{align}g^{\prime}(x) &= \left(\dfrac{1}{5}x^{-2} – \dfrac{1}{5}\right)^{\prime}\\ &= \dfrac{-2}{5}x^{-3}\\ &= \boxed{\dfrac{-2}{5x^3}}\end{align}$

In the example above, remember that the derivative of a constant is zero. This is why we no longer have $\dfrac{1}{5}$ in the answer.

Not bad right? For practice, you should try applying the quotient rule and verifying that you get the same answer.

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Continue your study of derivatives

Previous: The product rule
Next: The chain rule

The product rule for derivatives

Much of calculus and finding derivatives is about determining which rule applies to which case. The product rule, simply put, is applied when your function is the product of two other functions.

Formula for the product rule. For fg, write fg + fg, then take the derivative of the first and the derivative of the second to get f prime g plus f g prime.
In this guide, we will look at how to remember the product rule, how to recognize when it should be used, and finally, how to use it.

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Table of contents:

Remembering the product rule
Examples
Hint: Watch for shortcuts

Remembering the product rule

There is an easy trick to remembering this important rule: write the product out twice (adding the two terms), and then find the derivative of the first term in the first product and the derivative of the second term in the second product.
the-product-rule-steps

Examples

The easiest way to understand when this applies and how to use it is to look at some examples. In each case, pay special attention to how we identify that we are looking at a product of two functions.

Example

Find the derivative of the function.

$f(x) = x^4\ln(x)$

Solution

This function is the product of two simpler functions: $x^4$ and $\ln(x)$. Therefore, we can apply the product rule to find its derivative.

Write the product out twice, and put a prime on the first and a prime on the second:

$\left(f(x)\right)^{\prime} = \left(x^4\right)^{\prime}\ln(x) + x^4\left(\ln(x)\right)^{\prime}$

Take the derivatives using the rule for each function. Remember that for $x^4$, you will apply the power rule and that the derivative of $\ln(x)$ is $\dfrac{1}{x}$.

$\begin{align}\left(f(x)\right)^{\prime} &= \left(x^4\right)^{\prime}\ln(x) + x^4\left(\ln(x)\right)^{\prime}\\ &= \left(4x^3\right)\ln(x) + x^4\left(\dfrac{1}{x}\right)\end{align}$

Simplify, if possible.

$\begin{align}&= \left(4x^3\right)\ln(x) + x^3\\ &= \boxed{x^3\left(4\ln(x) + 1\right)}\end{align}$

Either of the last two lines can be used as a final answer, but the last one looks a little nicer and is probably going to be preferred by your teacher if you are currently taking calculus!

Let’s look at another example to make sure you got the basics down.

Example

Find the derivative of the function.

$y = 2xe^x$

This is the product of $2x$ and $e^x$, so we apply the product rule. Remember that the derivative of $2x$ is 2 and the derivative of $e^x$ is $e^x$.

$\begin{align}y^{\prime} &= \left(2x\right)^{\prime}e^x + 2x\left(e^x\right)^{\prime}\\ & = 2e^x + 2xe^x\\ &= \boxed{2e^x\left(1 + x\right)}\end{align}$

As you can see, with product rule problems, you are really just changing the derivative question into two simpler questions.

Hint: Watch for shortcuts

This hint could also be called “now that you know the product rule, don’t go blindly applying it”. To understand what that means, consider the following function:

$y = \left(x+4\right)\left(x+1\right)$

This is a product of $x+4$ and $x+1$, so if we want to find the derivative, we should use the product rule, right?

It’s true – you could use that. However, it is more work than recognizing that, by FOILing you get:

$y = \left(x+4\right)\left(x+1\right) = x^2+5x + 4$

Then by applying the power rule you have:

$y^\prime = \left(x^2+5x + 4\right)^{\prime} = 2x + 5$

This is the kind of thing you want to learn to notice. There are many problems where you can save yourself some calculus workby simplifying ahead of time. In the examples before, however, that wasn’t possible, and so the product rule was the best approach.

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Summary

The product rule is used to find the derivative of any function that is the product of two other functions. The quickest way to remember it is by thinking of the general pattern it follows: “write the product out twice, prime on 1st, prime on 2nd”.

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Previous: The power rule for derivatives

The derivative of a constant (a number)

The derivative of any constant (which is just a way of saying any number), is zero.

the-derivative-of-a-constant-or-number

This is easy enough to remember, but if you are a student currently taking calculus, you need to remember the many different forms a constant can take. First, let’s look at the more obvious cases.

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Example

Find the derivative of each function.

$\text{(a) } f(x) = 1$

$\text{(b) } g(x) = 20$

$\text{(c) } k(x) = -\dfrac{117}{91}$

Solution

No matter how cute we try to get with crazy fractions, one fact remains: each of these are constants. Therefore, the derivative of each is zero.

$\text{(a) } f^{\prime}(x) = \left(1\right)^{\prime} = 0$

$\text{(b) } g(x) = \left(20\right)^{\prime}=0$

$\text{(c) } k(x) = \left(-\dfrac{117}{91}\right)^{\prime}=0$

Constants in “disguise”

You learn about quite a few different types of constants in math. A couple that immediately come to mind are:

$e \approx 2.718$

$\pi \approx 3.142$

These are famous, but there are others that you have certainly worked with. Consider $\sqrt{2}$ or $\ln\left(5\right)$. Both of these are constants (if you aren’t sure, type them in your calculator – you will get the decimal equivalent) and so their derivatives are zero as well.

Example

Find the derivative of each of the following.

$\text{(a) } \pi^{3}$

$\text{(b) } \dfrac{\sqrt[3]{10}}{2}$

$\text{(c) } -(e-1)$

Solution

Again, each of these is a constant with derivative zero.

$\text{(a) } \left(\pi^{3}\right)^{\prime}=0$

$\text{(b) } \left(\dfrac{\sqrt[3]{10}}{2}\right)^{\prime}=0$

$\text{(c) } \left(-(e-1)\right)^{\prime}=0$

Don’t be fooled though. If we put an $x$ or other variable with any of these, they are no longer constants and the rules for finding their derivative is different. For example, $\left( e^x \right)^{\prime} = e^x$, not zero.

Why is the derivative of a constant zero?

One way of thinking about the derivative, is as the slope of a function at a given point. With functions like $f(x) = x^2$ (graphed below), the slope can change from point to point because the graph is curved.

graph-of-x^2

But what does the function look like if it is a constant function? Below is the graph of $f(x) = 2.5$.

graph-of-a-constant-function

This graph is a line, so the slope is the same at every point. Further, it is a horizontal line. The slope of any horizontal line is zero. Since the graph of any constant function is a horizontal line like this, the derivative is always zero.

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Summary

You will probably never have trouble finding the derivative of a constant when it is part of a polynomial or other function. But, be careful at paying attention to the different forms a constant may take, as professors and teachers love checking if you notice things like that. Further, you can use this easy idea to help you remember the concept of the derivative as the slope at a point – something that you will work with even when the derivatives are much more complicated.

The power rule for derivatives

Usually the first shortcut rule you study for finding derivatives is the power rule. The reason is that it is a simple rule to remember and it applies to all different kinds of functions. For a number n, the power rule states:

the-power-rule-for-derivatives

Let’s start with some really easy examples to see it in action.

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Example

Find the derivative of each function.

$\text{(a) } x^4$

$\text{(b) } x^{10}$

$\text{(c) } x^{546}$

Solution

For each of these, you can simply apply the power rule without any algebra at all. This means that you should bring the exponent out front, and then subtract 1 from the exponent.

$\text{(a) } \left(x^4\right)^{\prime} = 4x^3$

$\text{(b) } \left(x^{10}\right)^{\prime} = 10x^9$

$\text{(c) } \left(x^{546}\right)^{\prime} = 546x^{545}$

As you can see, it is all about remembering the pattern. Now, we will see how this pattern can be applied to more complicated examples.

Derivatives of polynomial functions

Recall that the derivative of a constant is always zero. So, the derivative of 5 is 0 while the derivative of 2,000 is also 0. Further, you can break the derivative up over addition/subtraction and multiplication by constants. Combining these ideas with the power rule allows us to use it for finding the derivative of any polynomial.

Example

Find the derivative of the function.
$y = 2x^4 – 5x^2 + 1$

Solution

With a little bit of practice, you will probably be able to write the derivative of this function down without thinking. Since it is our first example though, let’s write out every step. In the first step, we will break the derivative up over the addition and subtraction.

$\begin{align} y^{\prime} &= \left(2x^4 – 5x^2 + 1\right)^{\prime}\\ &= \left(2x^4\right)^{\prime} – \left(5x^2\right)^{\prime} + \left(1\right)^{\prime}\end{align}$

Now, factor out the coefficients:

$= 2\left(x^4\right)^{\prime} – 5\left(x^2\right)^{\prime} + \left(1\right)^{\prime}$

Apply the power rule for derivatives and the fact that the derivative of a constant is zero:

$= 2\left(4x^3\right) – 5\left(2x^1\right) + \left(0\right)$

Notice that once we applied the derivative rule, the prime went away. The correct notation keeps this until you apply a derivative rule. Now all we need to do is simplify to get our final answer.

$= \boxed{8x^3 – 10x}$

Let’s look at one more example without so much explanation to distract us.

Example

Find the derivative of the function.
$y = 5x^3 – 3x^2 + 10x – 8$

Solution

Apply the power rule, the rule for constants, and then simplify. Note that if $x$ doesn’t have an exponent written, it is assumed to be 1.

$\begin{align} y^{\prime} &= \left(5x^3 – 3x^2 + 10x – 8\right)^{\prime}\\ &= 5\left(3x^2\right) – 3\left(2x^1\right) + 10\left(x^0\right)- 0\end{align}$

Since $x$ was by itself, its derivative is $1x^0$. Normally, this isn’t written out however. Just remember that anything (other than zero) to the zero power is 1. So, $10\left(x^0\right) = 10(1) = 10$. Therefore, we can write the final answer as:

$= \boxed{15x^2 – 6x + 10}$

You may think this is all you can really do with the power rule. However, a couple of old algebra facts can help us apply this to a wider range of functions. We will look at two of those instances below.

Derivatives of functions with negative exponents

The power rule applies whether the exponent is positive or negative. But sometimes, a function that doesn’t have any exponents may be able to be rewritten so that it does, by using negative exponents. If this is the case, then we can apply the power rule to find the derivative. The main property we will use is:
$negative-exponents-to-fractions$
Let’s see what we can do with this property using an example!

Example

Find the derivative of the function.
$y = \dfrac{2}{x^4} – \dfrac{1}{x^2}$

Solution

Applying the rule for negative exponents, we can rewrite this function as:

$y = \dfrac{2}{x^4} – \dfrac{1}{x^2} = 2x^{-4} – x^{-2}$

Now that this is written with exponents, we can apply the power rule:

$\begin{align} y^{\prime} &= \left(2x^{-4} – x^{-2}\right)^{\prime}\\ &= 2\left(-4x^{-4-1}\right) – \left(-2x^{-2-1}\right)\end{align}$

Simplify to find the final answer:

$\begin{align} &= -8x^{-5} +2x^{-3}\\ &= \boxed{-\dfrac{8}{x^{5}} + \dfrac{2}{x^{3}}}\end{align}$

In the last step, notice that only the terms with the negative exponent were moved to the bottom of the fraction. The 8 didn’t have a negative exponent, so it stayed.

Derivatives of functions with radicals (square roots and other roots)

Another useful property from algebra is the following.
roots-to-exponents
Using this rule, we can take a function written with a root and find its derivative using the power rule.

Example

Find the derivative of the function.
$y = 4\sqrt{x} – 6\sqrt[3]{x^2}$

Solution

First, rewrite the function using algebra:

$y = 4\sqrt{x} – 6\sqrt[3]{x^2} = 4x^{\frac{1}{2}} – 6x^{\frac{2}{3}}$

Now apply the power rule:

$\begin{align} y^{\prime} &= \left(4x^{\frac{1}{2}} – 6x^{\frac{2}{3}}\right)^{\prime}\\ &= 4\left(\dfrac{1}{2}x^{\frac{1}{2}-1}\right) – 6\left(\dfrac{2}{3}x^{\frac{2}{3}-1}\right)\end{align}$

Simplify to get the final answer, remembering the rule for negative exponents:

$\begin{align} &= 2x^{-\frac{1}{2}} – 4x^{-\frac{1}{3}}\\ &= \dfrac{2}{x^{\frac{1}{2}}} – \dfrac{4}{x^{\frac{1}{3}}}\\ &= \dfrac{2}{\sqrt{x}} – \dfrac{4}{\sqrt[3]{x}}\end{align}$

In many classes, either of the last two lines can be written as your final answer. They are equivalent. However, your teacher or professor may have a preference, so always ask!

Summary

As a student studying calculus, you want the power rule to be second nature. It’ll be applied not only like this – on its own – but also as part of other rules such as the chain rule, the quotient rule, and the product rule. The better you understand it, the more you can focus on those more complicated ideas.

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Previous: The derivative of a constant

Next: The product rule

The power rule for integrals

The power rule for integrals allows us to find the indefinite (and later the definite) integrals of a variety of functions like polynomials, functions involving roots, and even some rational functions. If you can write it with an exponents, you probably can apply the power rule.

power-rule-integration

To apply the rule, simply take the exponent and add 1. Then, divide by that same value. Finally, don’t forget to add the constant C.

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Integrals of polynomials

Finding the integral of a polynomial involves applying the power rule, along with some other properties of integrals. Take a look at the example to see how.

Example

Find: $\displaystyle\int 2x^3 + 4x^2 \text{ dx}$

Solution

We will write out every step here so that you can see the process. After some practice, you will probably just write the answer down immediately.

First, remember that integrals can be broken up over addition/subtraction and multiplication by constants. Therefore:

$\begin{align} \displaystyle\int 2x^3 + 4x^2 \text{ dx} &= \displaystyle\int 2x^3\text{ dx} + \displaystyle\int 4x^2 \text{ dx}\\ &= 2\displaystyle\int x^3\text{ dx} + 4\displaystyle\int x^2 \text{ dx}\end{align}$

Now apply the power rule by adding 1 to each exponent, and then dividing by the same number. When you do this, the integral symbols are dropped since you have “taken the integral”.

$2\displaystyle\int x^3\text{ dx} + 4\displaystyle\int x^2 \text{ dx} = 2\left(\dfrac{x^{3+1}}{3+1}\right) + 4\left(\dfrac{x^{2+1}}{2+1}\right) + C$

Now, simplify the expression to find your final answer.

$\begin{align} &=2\left(\dfrac{x^{3+1}}{3+1}\right) + 4\left(\dfrac{x^{2+1}}{2+1}\right) + C\\ =& 2\left(\dfrac{x^{4}}{4}\right) + 4\left(\dfrac{x^{3}}{3}\right) + C\\ & = \bbox[border: 1px solid black; padding: 2px]{\dfrac{x^4}{2} + \dfrac{4x^3}{3} + C}\end{align}$

Now, let’s look at how this kind of integral would be with skipping some of the more straightforward steps.

Example

Find: $\displaystyle\int -3x^2 + x – 5 \text{ dx}$

Solution

This one is a little different. We have an $x$ by itself and a constant. For the $x$ by itself, remember that the exponent is 1. For the constant, remember that the integral of a constant is just the constant multiplied by the variable. For example, the integral of 2 with respect to $x$ is $2x$.

Applying the power rule:

$\displaystyle\int -3x^2 + x – 5 \text{ dx} = -3\left(\dfrac{x^3}{3}\right) + \dfrac{x^2}{2} – 5x + C$

Simplifying, the final answer is:

$ = \bbox[border: 1px solid black; padding: 2px]{-x^3 + \dfrac{x^2}{2} – 5x + C}$

Integrals of radical functions (functions with roots)

In order to apply this rule to this type of function, you must remember one very important idea from algebra.

integration-of-roots

So, if we can write the function using exponents then we can likely apply the power rule.

Example

Find: $\displaystyle\int \sqrt{x} + 4 \text{ dx}$

Before applying any calculus, you can rewrite the integral using the rule above. This way, we have exponents and can apply the power rule.

$\displaystyle\int \sqrt{x} + 4 \text{ dx} = \displaystyle\int {x}^{\frac{1}{2}} + 4 \text{ dx}$

Now, applying the power rule (and the rule for integrating constants):

$\displaystyle\int {x}^{\frac{1}{2}} + 4 \text{ dx} = \dfrac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} + 4x + C$

Simplify to get the final answer:

$\begin{align} &=\dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} + 4x + C\\ &= \bbox[border: 1px solid black; padding: 2px]{\dfrac{2}{3}x^{\frac{3}{2}} + 4x + C}\end{align}$

Usually, the final answer can be written using exponents like we did here or with roots. Your teacher or professor may have a preference, so make sure to ask!

Let’s work with one that is a little more messy with the fractions. As you will see, no matter how many fractions you are dealing with, the approach will stay the same.

Example

Find: $\displaystyle\int \dfrac{1}{2}\sqrt[3]{x} + 5\sqrt[4]{x^3} \text{ dx}$

Solution

Rewrite using algebra before you apply calculus rules so that you can use the power rule.

$\displaystyle\int \dfrac{1}{2}\sqrt[3]{x} + 5\sqrt[4]{x^3} \text{ dx}= \displaystyle\int \dfrac{1}{2}x^{\frac{1}{3}} + 5x^{\frac{3}{4}} \text{ dx}$

Apply the power rule:

$\displaystyle\int \dfrac{1}{2}x^{\frac{1}{3}} + 5x^{\frac{3}{4}} \text{ dx} = \dfrac{1}{2}\left(\dfrac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right) + 5\left(\dfrac{x^{\frac{3}{4}+1}}{\frac{3}{4}+1}\right) +C$

Simplify to get the final answer:

$\begin{align} &= \dfrac{1}{2}\left(\dfrac{x^{\frac{4}{3}}}{\frac{4}{3}}\right) + 5\left(\dfrac{x^{\frac{7}{4}}}{\frac{7}{4}}\right) +C\\ &= \dfrac{1}{2}\left(\dfrac{3}{4}{x^{\frac{4}{3}}}\right) + 5\left(\dfrac{4}{7}x^{\frac{7}{4}}\right) +C\\ &= \bbox[border: 1px solid black; padding: 2px]{\dfrac{3}{8}x^{\frac{4}{3}} + \dfrac{20}{7}x^{\frac{7}{4}} +C}\end{align}$

Did you notice that most of the work was with algebra? This is true of most calculus problems. The calculus part is straightforward while the algebra requires you to be very careful and makes up most of the problem.

Integrals with negative exponents

One more old algebra rule will let us use the power rule to find even more integrals.
power-rule-integrals-negative-exponents
Let’s see how we can apply it!

Example

Find: $\displaystyle\int \dfrac{3}{x^5} – \dfrac{1}{4x^2} \text{ dx}$

Use the rule above and rewrite this integral with exponents. Pay special attention to what terms the exponent applies to.

$\displaystyle\int \dfrac{3}{x^5} – \dfrac{1}{4x^2} \text{ dx} = \displaystyle\int 3x^{-5} – \dfrac{1}{4}x^{-2} \text{ dx}$

Apply the power rule:

$\displaystyle\int 3x^{-5} – \dfrac{1}{4}x^{-2} \text{ dx} = 3\left(\dfrac{x^{-5+1}}{-5+1}\right) – \dfrac{1}{4}\left(\dfrac{x^{-2+1}}{-2+1}\right) + C$

Simplify to get the final answer:

$\begin{align} &= 3\left(\dfrac{x^{-4}}{-4}\right) – \dfrac{1}{4}\left(\dfrac{x^{-1}}{-1}\right) + C\\ &= -\dfrac{3}{4}x^{-4} + \dfrac{1}{4}x^{-1} + C\\ &= -\dfrac{3}{4}\left(\dfrac{1}{x^4}\right) + \dfrac{1}{4}\left(\dfrac{1}{x}\right) + C\\ &= \bbox[border: 1px solid black; padding: 2px]{-\dfrac{3}{4x^4} + \dfrac{1}{4x} + C}\end{align}$

Note that this only works when the exponent is not –1. If you tried to apply the power rule here, you would end up dividing by zero. There is a different rule for dealing with functions like $\dfrac{1}{x}$.

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Summary

As you have seen, the power rule can be used to find simple integrals, but also much more complicated integrals. The general strategy is always the same – if you don’t already have exponents, see if you can write the function using exponents. Then, apply the power rule and simplify.

Oh yeah, and don’t forget to add C!

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The chain rule

The chain rule is a method for finding the derivative of composite functions, or functions that are made by combining one or more functions. An example of one of these types of functions is $f(x) = (1 + x)^2$ which is formed by taking the function $1+x$ and plugging it into the function $x^2$. A surprising number of functions can be thought of as composite and the chain rule can be applied to all of them.

Table of Contents

How the formula for the chain rule works
Examples of applying the chain rule
Summary

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How the formula for the chain rule works

The chain rule says that if $h$ and $g$ are functions and $f(x) = g(h(x))$, then

The chain rule formula. For f(x) = g(h(x)), the derivative is g prime of h of x times h prime of x.

This looks complicated, so let’s break it down. The main function $f(x)$ is formed by plugging $h(x)$ into the function $g$. You can think of $g$ as the “outside function” and $h$ as the “inside function”. Using the chain rule, if you want to find the derivative of the main function $f(x)$, you can do this by taking the derivative of the outside function $g$ and then multiplying it by the derivative of the inside function $h$. In other words, you are finding the derivative of $f(x)$ by finding the derivative of its pieces.

Examples using the chain rule

As we apply the chain rule, we will always focus on figuring out what the “outside” and “inside” functions are first. From there, it is just about going along with the formula.

Example

Find the derivative of $f(x) = (3x + 1)^5$.

Solution

In this example, there is a function $3x+1$ that is being taken to the 5th power. So, there are two pieces: the $3x + 1$ (the inside function) and taking that to the 5th power (the outside function). You know by the power rule, that the derivative of $x^5$ is $5x^4$. So, cover up that $3x + 1$, and pretend it is an $x$ for a minute. The only deal is, you will have to pay a penalty. Since it was actually not just an $x$, you will have to multiply by the derivative of the $3x+1$.

Chain rule example. The derivative of left parentheses 3 x plus 1 right parentheses to the fifth power is 5 left parentheses 3 x plus 1 right parentheses to the 4th power times the derivative of 3 x plus 1, which is 3.

Simplify to get the final answer:

$f'(x) = \boxed{5(3x+1)^4(3) = 15(3x+1)^4}$

Example

Find the derivative of $f(x)=\ln(x^2-1)$.

Solution

The same idea will work here. Normally, if it was just $\ln(x)$, you would say the derivative is $\dfrac{1}{x}$. However, there is something there other than $x$ (the inside function). So, cover it up and take the derivative anyway. Just don’t forget to multiply by the derivative of the inside function after you are done.

Second example of finding the derivative using the chain rule. The derivative of ln of x squared minus 1 is one over x squared minus 1 times the derivative of x squared minus 1.

Now you can simplify to get the final answer:

$f'(x)= \boxed{\dfrac{2x}{x^2-1}}$

If you need to review taking the derivative of ln(x), see this lesson: https://www.mathbootcamps.com/derivative-natural-log-lnx/

Summary

Composite functions come in all kinds of forms so you must learn to look at functions differently. When trying to decide if the chain rule makes sense for a particular problem, pay attention to functions that have something more complicated than the usual $x$. Some examples are $e^{5x}$, $\cos(9x^2)$, and $\dfrac{1}{x^2-2x+1}$. All of these are composite functions and for each of these, the chain rule would be the best approach to finding the derivative.

Reader Question: Calculus Without Trigonometry?

Woohoo I LOVE email! My love of email may go so far as to be distracting, but that is a completely different topic. Just make a note: If you ever have any questions about doing well in math, send ’em my way! You can get a hold of me through jerimi@mathbootcamps.com. (and NO I won’t do your math homework for you)

Last week, a reader asked if it was possible for him to do well in his calculus course even though he had never studied any trigonometry. According to him, his algebra skills are solid and so far he has been able to manage the course work.

There is a reason why a lot of high schools and colleges combine trig with algebra in a course like precalculus. If you are willing to put in a little bit of outside effort, many topics in trig are easy to pick up and there are really only a few key skills/ideas. Don’t get me wrong, there is a whole lot of memorization and things like solving a trig equation WILL come up in a calculus course. But, for someone who is able to learn math on their own, picking it up along the way is possible.

Will it be easy? No way. I warned our friend that instead of being able to focus on the new calculus topics by themselves like everyone else, his studying time will also be filled with learning the trig. Calculus is already a challenging course by itself!

For anyone else considering this, I recommend against it unless you are the type who generally can learn math on their own, and is willing to work through a book like schaum’s outlines at the same time as working through your calculus problems. Even then, it will be a tough road and you might not get the grade you would have if you had been able to focus on the calculus alone. It is all about how much time you are willing to put in and how well you use all the resources available to you.

Calculus Problem of the Week November 18, 2011

Sometimes easy and sometimes hard, our calculus problem of the week could come from any calculus topic. If you really want to get better at calculus, following these problems is a great way to make yourself practice! Past calculus problems of the week.

This week’s problem:
Earlier today, I managed to find a really amazing new (to me) math blog. Not only that, there was this (insanely) interesting and creative calculus problem! So, of course, I must share! The problem is: (quoted)

“You’re in a museum and you’re looking at a painting which is hung above eye level. (There is a specific painting which is hung high in the entrance room at the Brooklyn Museum that I think of with this problem.) You are standing some distance away from it. The question is: what is the largest angle x that you can get as you walk forwards and backwards? (See diagram below for setup.)”

Head over this way to see the answer and more!