# Chain rule problem

• Feb 11th 2010, 12:41 AM
Kakariki
Chain rule problem
Hey helpful member of this forum, I am currently in a calculus and vectors self learning course, and I need help with a problem. ***As I wrote this thread I was doing all the math again. I found that my problem is my finding of the derivative of $y = u(u^2 + 3)^3$. I kept the rest of my solution up just incase this is also a problem. My main concern, however, is my bold statement showing where the main problem (I think) is.***

Question: Use the chain rule to find dy/dx at the indicated value of x.
$y = u(u^2 + 3)^3, u = (x + 3)^2, x = -2$

My solution:
Basically how the text explains the chain rule is: "If f and g are functions having derivatives, then the composite function $h(x) = f(g(x))$ has a derivative given by $h'(x) = f'(g(x))g'(x)$"

So with this information this is how my solution looks:
First I expanded $u = (x+3)^2$:
$u = x^2 + 6x + 9$

Which makes it easier for me to find the derivative (just me likely):
$du/dx = (1)(2)x^{2-1} + (1)(6)x^{1-1}$
$= 2x + 6$

I believe this is where the problem lies!!! So help in finding the derivative here would be awesome!!!
Then I found the derivative of $y = u(u^2 + 3)^3$:
$dy/du = (u)(3)(u^2 + 3)^{3-1}((1)(2)u^{2-1})$
$dy/du = 3u(u^2 + 3)^2(2u)$
$dy/du = 6u^2(u^2 + 3)^2$*
*According to wolfram alpha, this derivative is incorrect. I do not understand why. The solution on wolfram alpha when I input $y = u(u^2 + 3)^2$ is $(3 + u^2)^2 (3 + 7u^2)$. I do not understand how they come to this answer.* Link: http://www.wolframalpha.com/input/?i=y+%3D+u%28u^2+%2B+3%29^3

Assuming the above is correct, subbing $u = (x + 3)^2$ into $y = u(u^2 + 3)^3$ should *assuming the answer in the book is correct* come to 320 when I sub in -2 for x. So:
$(6((x)^2 + 6(x) + 9)^2)((((x)^2 + 6(x) + 9)^2) +3)^2$
Then I sub in x = -2:
$(6((-2)^2 + 6(-2) + 9)^2)((((-2)^2 + 6(-2) + 9)^2) +3)^2$
$= (6((1)^2))(4)^2$
$= (6)(16)$
$= 96$ Then multiply by 2x + 6:
$(96)(2)$
$= 192$

When I use the derivative shown on wolfram alpha I do get 320. *I tried this before but did not get correct answer, just tried again now and did!!!* How do I find the derivative shown on wolfram alpha?
$(3 + (x^2 + 6x + 9)^2)^2 * (3 + 7(x^2 + 6x + 9)^2)$
Sub in x = -2:
$( 3 + ((-2)^2 + 6(-2) + 9)^2)^2 * (3 + 7((-2)^2 + 6(-2) + 9)^2)$
$= ((3 + 1)^2)(3 + 7(1))$
$= (16)(10)$
$= (160)(2)$ *multiply by (2x + 6) with x = -2 subbed in*
$= 320$

- Thanks for looking at my thread!
• Feb 11th 2010, 01:06 AM
Calculus26
another way of thinking of the chain rule is

If y = y (u) and u = u(x)

Then dy/dx = (dy/du)(du/dx)

For y= u (u^2 + 3) ^3 where u = (x+3)^2

dy/du = (u^2 + 3) ^3 + 6u^2 (u^2 + 3) ^2 (Product rule)

dy/du = [(u^2+3)^2] [7u^2 +3]

du/dx = 2(x+3)

dy/dx = 2[(u^2+3)^2] [7u^2 +3](x+3)

Normally we would now convert u to (x+3)^2 but we only need the derivative when x= -2 . Note if x= -2 then u = 1

substitute into the above eqn.