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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:
Take the derivative of:
[tex] f(x) = \frac{sin(x^2)}{ln sinx}[/tex]
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]
[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.
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.