# Finding the derivative of $\ln(\sin(x))$ using first principles.

Let $y=f(x)=\ln(\sin(x))$

$f(x+h)=\ln(\sin(x+h))$

$$\frac{d}{dx}(y)=\frac{d}{dx}(f(x))=\lim_{h \to0}\frac{f(x+h)-f(x)}{h}=\lim_{h \to0}\frac{\ln(\sin(x+h))-\ln(\sin(x))}{h}$$ $$=\lim_{h \to0}\frac{\ln\left(\frac{\sin(x+h)}{\sin(x)}\right)}{h} =\lim_{h \to0}\frac{\ln\left(\frac{\sin(x)\cos(h)+\cos(x)\sin(h)}{\sin (x)}\right)}{h}$$ $$=\lim_{h \to0}\frac{\ln (\cos(h)+\cot(x)\sin(h))}{h}=\lim_{h \to0}\frac{\ln( \cos(h)(1+\cot(x)\tan(h)))}{h}$$

• Using $\log x \sim x-1$ if $x\to 1$ – Rafael Gonzalez Lopez Feb 15 '18 at 7:36

$$\ln\dfrac{1+\dfrac{\sin(x+h)-\sin x}{\sin x}}h$$

$$=\ln\dfrac{\left(1+\dfrac{\sin(x+h)-\sin x}{\sin x}\right)}{\dfrac{\sin(x+h)-\sin x}{\sin x}}\cdot\dfrac{\sin(x+h)-\sin x}{h\sin x}$$

Now $\lim_{u\to0}\dfrac{\ln(1+u)}u=1$

and $$\lim_{h\to0}\dfrac{\sin(x+h)-\sin x}h=\lim_{h\to0}\dfrac{\sin\dfrac h2\cos\left(x+\dfrac h2\right)}{\dfrac h2}=?$$

• Correct me if I'm wrong, but the derivative of the function $\ln(\sin x)$ derived using chain rule is $\frac{\cos x}{\sin x}$, right? But by using the property that $\lim_{h\to0}\frac{\sin h}{h} = 1$, the solution of the limit in your answer is $\cos x$. – Gil Keidar Feb 17 '18 at 6:00
• @GilKeidar, Have you noticed $h\sin x$ somewhere in the post? – lab bhattacharjee Feb 17 '18 at 10:39

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.

• Use L'Hopital's rule to eliminate the variable in the denominator, or alternatively, for your denominator, find a function y = f(x) = not zero. Either method will help find the limit of this function. – Gwopmeat Feb 19 '18 at 6:43

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