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Solve for $\frac{dy}{dx}$ if $y=\ln(\sin x+\ln x)$.

I know how to solve for integrals involving $du$ and $u$, but how do I do this type of problem (I think it's the opposite of the integral problem)?

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@5space Thanks for the edit. I still don't know how to write mathematical equations here in math.stackexchange.. – Gannicus May 15 '13 at 4:24
check this link out:… As well, you may like to look at a quick start guide to LaTeX. – user59083 May 15 '13 at 4:31
Oh, so it's LaTeX formatting? I've tried LaTeX before, but I didn't know LaTeX applies here, or anywhere outside of LaTeX editors. – Gannicus May 15 '13 at 4:45
up vote 7 down vote accepted

Use the "good old" chain rule: remember that?

$$y = \ln(f(x)) \iff y' = \dfrac{f'(x)}{f(x)}$$

$$y = \ln(\sin x + \ln x) \implies y' = \dfrac{\left(\cos x + \frac 1x\right)}{\sin x + \ln x}$$

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I doubt myself so much that this was my answer when I solved the problem earlier, but I wasn't too confident about it. Thanks for confirming! – Gannicus May 15 '13 at 4:23
@amWhy: still can't shut down I see! :-) =1 – Amzoti May 15 '13 at 4:25
I completely understand "self-doubt" or "second-guessing"; I've been there! – amWhy May 15 '13 at 4:25
Hearing that from someone who reaps 200+ reputation here a day, I am inclined to doubt that :p Haha! – Gannicus May 15 '13 at 4:33

You use the chain rule, which says $[f(g(x))]'=f'(g(x))g'(x)$

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