# Calculus made easy by Thompson: exercise vii: 1

1. Sep 4, 2013

### tridianprime

Hello, I require some assistance with this problem. It reads:

If, u=1/2x^3; v=3(u+u^2); w=1/v^2;

Find dw/dx.

It's clear that you should use the chain rule so I begin by calculating the derivative for each one.

I get v^(-2); 6u+3; and 3/2x^2.

I then have to multiply them but this is where the problem comes. The answer I derived was (3x^(2)-12u-6)/2v^3

This doesn't seem right and the answer say it isn't at the back of book.

Is the problem the multiplication?

2. Sep 4, 2013

v-2 = what?
6u+3 = what?
etc.

3. Sep 4, 2013

### tridianprime

In order of appearance: dw/dv then dv/du then du/dx.

4. Sep 4, 2013

### HallsofIvy

Staff Emeritus
The first is incorrect. $w= 1/v^2= v^{-2}$ itself. The derivative is $w'= -2v^{-3}$
but apparently that is a typo since you have $v^3$ below.
(u= (1/2)x^3, not 1/(2x^3)?)

Specifically, you have $(-2v^{-3})(6u+ 3)(3/2)x^2$. With [/itex]u= (1/2)x^3[/itex], $6u+ 3= 3x^3+ 3$ so this is -9(x^5+ 3x^2)/v^3

What is the answer in the book? Do they have x, u, and v or have the converted everything to x?

5. Sep 4, 2013

### tridianprime

Everything is x.

6. Sep 4, 2013