# Homework Help: Chain rule problem

1. Oct 11, 2005

### roe

I'm thoroughly confused as to how and work this problem. I thought I had an ok understanding of the chain rule when I started the section's homework, but this question has me ready to gorge out my eyeballs!

The Problem:
---------------
Find:dy/dx at x = 2
Given: y = (s+3)^2, s = sqrt(t-3), t = x^2

OK, So from what I understand, in this case dy/dx can be found using the chain rule, which is:

dy/dx = dy/ds * ds/dt * dt/dx

dt/dx is easy, and is just 2x. sqrt(t-3) looked trickier to me at first, but I'm getting it to be 1 (sqrt(t-3)' = sqrt(1-0) = 1, correct?). I've worked out (s+3)^2 two different ways, getting the same answer both times:

((s+3)(s+3))' = (s^2 + 6s + 9)' = 2s + 6 = 2(s + 3)
--or--
say y = h(s), f(s) = s^2, g(s) = (s+3)
then f'(s) = 2s and g'(s) = 1
h'(s) = f'(g(s))*g'(s) = f'(s+3) * 1 = 2(s+3) * 1 = 2(s + 3)

So I'm fairly certain that dy/ds is 2(s + 3).

Now I know that dy/ds is 2(s+3), ds/dt is 1, and dt/dx is 2x. Now I can multiply:

2(s+3) * (1) * (2x) ----> and put s+3 in terms of x by substituting:

2(sqrt(t-3) + 3) * (1) * (2x) -----> 2(sqrt(x^2 - 3) + 3) * (1) * (2x)

And finally, set x == 2:

2(sqrt(2^2 - 3) + 3) * (4) = 2(sqrt(1) + 3) * (4) = 2(1 + 3) * 4 =

2(4) * 4) = 8 * 4 = 32! The book's answer is 16, so I'm making a mistake somewhere but I'm having difficulty figuring out where. Any help is appreciated!

2. Oct 11, 2005

### SN1987a

Not correct actually. The chain rule applies here and tells you to derivate the outer function first and then multiply by the derivative of the inner function. The outer function in your case is the square root, while the inner function is t-3. Therefore (sqrt(t-3))' = 1/(2*sqrt(t-3)). This is very much the same to what you did to find dy/ds (2nd method), which is correct by the way.

All the rest is correct, and if you redo your calculations with the right ds/dt, you should indeed get 16.

3. Oct 11, 2005

### roe

Aha! So, if I say:

h(x) = sqrt(t-3) = (t-3)^.5

f(t) = t^.5
g(t) = t-3

f'(t) = .5(t)^-.5
g'(t) = 1

so:

h'(t) = f'(g(t)) * g'(t) -------->
h'(t) = f'(t-3) * 1 -------->

h'(t) = 1/2 * (t-3)^(-1/2) ----> h'(t) = 1/2 * 1 / sqrt(t-3) ------->

h'(t) = 1 / 2 * sqrt(t-3), then I solved for x = 2 and got 16. Thank you so much for making this clear!