Finding the derivative of $\ln(\sin(x))$ using first principles.
Let $y=f(x)=\ln(\sin(x))$
$f(x+h)=\ln(\sin(x+h))$
$$\frac{d}{dx}(y)=\frac{d}{dx}(f(x))=\lim_{h \to0}\frac{f(x+h)-f(x)}{h}=\lim_{h \to0}\frac{\ln(\sin(x+h))-\ln(\sin(x))}{h}$$ $$=\lim_{h \to0}\frac{\ln\left(\frac{\sin(x+h)}{\sin(x)}\right)}{h} =\lim_{h \to0}\frac{\ln\left(\frac{\sin(x)\cos(h)+\cos(x)\sin(h)}{\sin (x)}\right)}{h}$$ $$=\lim_{h \to0}\frac{\ln (\cos(h)+\cot(x)\sin(h))}{h}=\lim_{h \to0}\frac{\ln( \cos(h)(1+\cot(x)\tan(h)))}{h}$$
I am stuck at this step. Please help.
$\endgroup$ 13 Answers
$\begingroup$$$\ln\dfrac{1+\dfrac{\sin(x+h)-\sin x}{\sin x}}h$$
$$=\ln\dfrac{\left(1+\dfrac{\sin(x+h)-\sin x}{\sin x}\right)}{\dfrac{\sin(x+h)-\sin x}{\sin x}}\cdot\dfrac{\sin(x+h)-\sin x}{h\sin x}$$
Now $\lim_{u\to0}\dfrac{\ln(1+u)}u=1$
and $$\lim_{h\to0}\dfrac{\sin(x+h)-\sin x}h=\lim_{h\to0}\dfrac{\sin\dfrac h2\cos\left(x+\dfrac h2\right)}{\dfrac h2}=?$$
$\endgroup$ 2 $\begingroup$Your answer is correct lab, but the formal definition of a derivative is the ugliness provided in the picture above, and not the nice and neat cot(x) you described. Both methods, the chain rule and formal definition of a derivative, will always provide the same result - a derivative of a univariate function.
$\endgroup$ 1 $\begingroup$Instead of dividing the top part by $\sin x$, take $\sin h$ out. So you have $$\ln(\cot h+\cot x)\over h$$ Now use L'Hôpital
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