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Need some help with a derivative and the chain rule... 
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#1
Mar410, 02:30 PM

P: 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: [tex] f(x) = \frac{sin(x^2)}{ln sinx}[/tex] 2. Relevant equations Here's the formula I used (and always do) for the chain rule: [tex] \frac{d}{dx}[f(g(x))] = f'(g(x))*g'(x) [/tex] 3. The attempt at a solution [tex] f'(x) = \frac{ln sinx(cos(x^2)2x)  sin(x^2)\frac{1}{sinx}sinx cosx}{[ln sinx]^2}[/tex] So, I've done this simple derivative problem 3 times and for some reason I keep getting this answer. EMailed 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
Mar410, 02:36 PM

P: 610

Why do you have a sinx there? sinx^{2} comes from the original function, cosx/sinx comes from differentiating ln(sinx).



#3
Mar410, 02:43 PM

P: 154

I'm thinking [tex]g(x) = sinx[/tex] so by the chain rule the bottom [tex]lnsinx = \frac{1}{sinx}sinxcosx[/tex]
And you and my prof are saying that [tex]lnsinx = \frac{1}{sinx}cosx[/tex] But what happens to the [tex]g(x)[/tex] 


#4
Mar410, 02:45 PM

P: 610

Need some help with a derivative and the chain rule...
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
Mar410, 03:07 PM

P: 154

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



#6
Mar410, 03:09 PM

P: 154




#7
Mar410, 03:13 PM

P: 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
Mar410, 03:16 PM

P: 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
Mar410, 03:22 PM

P: 610

No problem! :) Peace



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