﻿ Calculus Problem of the Week July 4, 2011 - MathBootCamps

# Calculus Problem of the Week July 4, 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
Find the derivative of $\dfrac{x^4}{x}$ using THREE DIFFERENT METHODS.
See the solution.

It may seem strange to find this derivative in three different ways, but it allows you to practice thinking about problems differently. If you can recognize different ways to approach a given problem, you can more easily decide which is the best. (it also gives you a chance to really practice your algebra skills!)

I will go through three of the more “obvious” approaches. If you found another way – please post!

Method 1: Simplify first.

Using the laws of exponents, $\dfrac{x^4}{x}=x^3$. Therefore the derivative is $3x^2$ (by the power rule)

Method 2: The quotient rule. $(\dfrac{x^4}{x})'=\dfrac{(x^4)'(x)-(x^4)(x)'}{x^2}=\dfrac{(4x^3)(x)-x^4}{x^2}=\dfrac{4x^4-x^4}{x^2}=\dfrac{3x^4}{x^2}=3x^2$. It wasn’t pretty but it worked!

Method 3: The product rule.

First rewrite $\dfrac{x^4}{x}$ as a product: $(x^4)(x^{-1})$. Now: $((x^4)(x^{-1}))'=(x^4)'(x^{-1})+(x^4)(x^-1)'=(4x^3)(x^{-1})+(x^4)(-x^{-2})=\dfrac{4x^3}{x}-\dfrac{x^4}{x^2}=4x^2-x^2=3x^2$.

PS you may have noticed that I used a slightly different way of writing out the product and quotient rules. While it is equivalent to the form you find in textbooks, I have always found this way easier to remember for myself and students! 