# Need some help with a derivative and the chain rule...

1. ### sEsposito

154
Okay so I'm doing chain rule work to go over the stuff from calc 1 before I take a departmental exam and I've run into this problem:

1. The problem statement, all variables and given/known data
Take the derivative of:

$$f(x) = \frac{sin(x^2)}{ln sinx}$$

2. Relevant equations
Here's the formula I used (and always do) for the chain rule:

$$\frac{d}{dx}[f(g(x))] = f'(g(x))*g'(x)$$

3. The attempt at a solution

$$f'(x) = \frac{ln sinx(cos(x^2)2x) - sin(x^2)\frac{1}{sinx}sinx cosx}{[ln sinx]^2}$$

So, I've done this simple derivative problem 3 times and for some reason I keep getting this answer. E-Mailed my Prof and he said that the sinx right after the 1/sinx is not suposed to be there, but I cannot see why...Can anyone verify the correct answer for me just to see what I'm not grasping here. I'm very frustrated.

2. ### VeeEight

610
Why do you have a sinx there? sinx2 comes from the original function, cosx/sinx comes from differentiating ln(sinx).

3. ### sEsposito

154
I'm thinking $$g(x) = sinx$$ so by the chain rule the bottom $$lnsinx = \frac{1}{sinx}sinxcosx$$

And you and my prof are saying that $$lnsinx = \frac{1}{sinx}cosx$$

But what happens to the $$g(x)$$

4. ### VeeEight

610
D [ln(sinx)] = (1/sinx)cosx.
(where D is the derivative)

The general formula you are using is D [a(x)/b(x)] = [a'(x)b(x) - b'(x)a(x)] / [b(x)]2
You have most the terms right, it is the b'(x)a(x) term that is incorrect.

5. ### sEsposito

154
But why? If the rule is $$f'(x) = f'(g(x))*g'(x)$$ than what happens to the $$g(x)$$ which in this case $$= sinx$$

6. ### sEsposito

154
That's the quotient rule, but I'm talking about the chain rule that needs to be used to get $$f'(x)$$

7. ### VeeEight

610
Show step by step where the problem is.
If you are trying to differentiate ln(sinx), take the g(x)=sinx. then D [ln(g(x))] = 1/g(x) * g'(x). What is g'(x) here? It is cosx. Thus D [ln(g(x))] = 1/g(x) * g'(x) = (1/sinx)cosx

8. ### sEsposito

154
I see it now! You're right. I can't thank you enough.. This was driving me crazy. Seriously, thank you so much.

9. ### VeeEight

610
No problem! :) Peace