Calculus made easy by Thompson: exercise vii: 1

[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?

 

[Mentallic]

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.

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

 

[tridianprime]

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

 

[HallsofIvy]

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.

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)?)

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

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

Is the problem the multiplication?

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

 

[tridianprime]

Everything is x.

 

[tridianprime]

It reads: -(3x^2(3+3x^3))/(27((1/2)x^3 + (1/4)x^6)^3)