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!(adsbygoogle = window.adsbygoogle || []).push({});

The Problem:

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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!

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# Homework Help: Chain rule problem

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