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I thought the derivative of $\ln(u)$ was $\frac{u'}{u}$. So why does the derivative of $\ln(\sin x)$ = $\frac{1}{\sin x}(\cos x)$ instead of $\frac{\cos x}{\sin x}(\cos x)$?

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    If $u = \sin x$, then $\frac {u'}u = \frac{\cos x}{\sin x} = \cot x$. Why did the $\cos x$ come in brackets again? – 2017-01-12
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    From the chain rule is where I got the $(cosx)$ after. – 2017-01-12
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    I'm confused as to why $u' = 1$ instead of $\frac{d}{dx}(sinx)$ – 2017-01-12
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    Which chain rule? Apply the chain rule properly: let $f(x) = \sin x$ and $g(x) = \ln x$. We are then dealing with $(g \circ f)'(x) = g'(f(x)) \times f'(x) = \frac{1}{\sin x} \times \cos x$, which is the answer. So what's wrong with this working, or what don't you understand here? Ok, so I've seen your confusion. Wait for the next comment. – 2017-01-12
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    If $g(x)=\ln x$, then $g'(x) = \frac 1x$, so that's why the $1$ is on the numerator. The function $\frac 1x$ is applied on the argument $\sin x$ to give $\frac 1{\sin x}$, so no further derivative of $\sin x$ is required. – 2017-01-12
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    Okay I see my mistake thank you very much for this. – 2017-01-12
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    That's great, and you are welcome. $+1$ for your question. – 2017-01-12

1 Answers 1

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Some points I want to highlight:

$$\frac{d}{dx}(\ln u)$$ $$=\frac1u\cdot \frac{d}{dx}(u)$$

This is the result that your first line reports.

Now for $\ln (\sin x)$, observe that $u=\sin x$ and $\frac{d}{dx}(u)=\cos x$.

So from the aforementioned result,

$$\frac{d}{dx}(\ln \sin x)$$ $$=\frac{1}{\sin x}\cdot \frac{d}{dx}(\sin x)$$ $$=\frac{1}{\sin x}\cdot \cos x$$ $$=\cot x$$

Hence, $\boxed{\frac{d}{dx}(\ln \sin x)=\cot x}$.

Hope this helps you.