Need some help with a derivative and the chain rule

  • Thread starter sEsposito
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  • #1
sEsposito
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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:

[tex] f(x) = \frac{sin(x^2)}{ln sinx}[/tex]



Homework 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]



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. 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.
 

Answers and Replies

  • #2
VeeEight
614
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Why do you have a sinx there? sinx2 comes from the original function, cosx/sinx comes from differentiating ln(sinx).
 
  • #3
sEsposito
154
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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
VeeEight
614
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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
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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
sEsposito
154
0
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 [tex]f'(x)[/tex]
 
  • #7
VeeEight
614
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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
0
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
614
0
No problem! :) Peace
 

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