# 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 \$\displaystyle 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.
\$\displaystyle 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 \$\displaystyle h(x) = f(g(x)) \$ has a derivative given by \$\displaystyle h'(x) = f'(g(x))g'(x) \$"

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

Which makes it easier for me to find the derivative (just me likely):
\$\displaystyle du/dx = (1)(2)x^{2-1} + (1)(6)x^{1-1} \$
\$\displaystyle = 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 \$\displaystyle y = u(u^2 + 3)^3 \$:
\$\displaystyle dy/du = (u)(3)(u^2 + 3)^{3-1}((1)(2)u^{2-1}) \$
\$\displaystyle dy/du = 3u(u^2 + 3)^2(2u) \$
\$\displaystyle 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 \$\displaystyle y = u(u^2 + 3)^2 \$ is \$\displaystyle (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 \$\displaystyle u = (x + 3)^2 \$ into \$\displaystyle 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:
\$\displaystyle (6((x)^2 + 6(x) + 9)^2)((((x)^2 + 6(x) + 9)^2) +3)^2 \$
Then I sub in x = -2:
\$\displaystyle (6((-2)^2 + 6(-2) + 9)^2)((((-2)^2 + 6(-2) + 9)^2) +3)^2 \$
\$\displaystyle = (6((1)^2))(4)^2 \$
\$\displaystyle = (6)(16) \$
\$\displaystyle = 96 \$ Then multiply by 2x + 6:
\$\displaystyle (96)(2) \$
\$\displaystyle = 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?
\$\displaystyle (3 + (x^2 + 6x + 9)^2)^2 * (3 + 7(x^2 + 6x + 9)^2) \$
Sub in x = -2:
\$\displaystyle ( 3 + ((-2)^2 + 6(-2) + 9)^2)^2 * (3 + 7((-2)^2 + 6(-2) + 9)^2) \$
\$\displaystyle = ((3 + 1)^2)(3 + 7(1)) \$
\$\displaystyle = (16)(10) \$
\$\displaystyle = (160)(2) \$ *multiply by (2x + 6) with x = -2 subbed in*
\$\displaystyle = 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.