Need some help with a derivative and the chain rule

[sEsposito]

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:

Homework Statement

Take the derivative of:

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

Homework Equations

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

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

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.

 

[VeeEight]

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

 

[sEsposito]

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 $$g(x)$$

 

[VeeEight]

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.

 

[sEsposito]

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

 

[sEsposito]

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.

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

 

[VeeEight]

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

 

[sEsposito]

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

 

[VeeEight]

No problem! :) Peace