Question

Differentiate with respect to $x$ from first principles $f\left(x\right)=\log \sin x$

Answer 1

$f\left(x\right)=\log \sin x$.
Then, $f(x+h)=\log \sin (x+h)$
$\Rightarrow \frac{d}{dx}f\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow \frac{d}{dx}f\left(x\right)=\underset{h\to 0}{lim}\frac{\log \sin (x+h)-\log \sin \left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{\log \left(\frac{\sin (x+h)}{\sin x}\right)}{h}$
$=\underset{h\to 0}{lim}\frac{\log (1+\frac{\sin (x+h)}{\sin x}-1)}{h}$
$=\underset{h\to 0}{lim}\frac{\log (1+\frac{\sin (x+h)-\sin x}{\sin x})}{h}$
$=\underset{h\to 0}{lim}\frac{\log (1+\frac{\sin (x+h)-\sin x}{\sin x})}{h\left(\frac{\sin (x+h)-\sin x}{\sin x}\right)}\times \left[\frac{\sin (x+h)-\sin x}{\sin x}\right]$

$=\underset{h\to 0}{lim}\frac{\log (1+\frac{\sin (x+h)-\sin x}{\sin x})}{\left(\frac{\sin (x+h)-\sin x}{\sin x}\right)}\times [\frac{\sin (x+h)-\sin x}{h}\times \frac{1}{\sin x}]$

$=\underset{h\to 0}{lim}\frac{\log (1+\frac{\sin (x+h)-\sin x}{\sin x})}{\left(\frac{\sin (x+h)-\sin x}{\sin x}\right)}\times \underset{h\to 0}{lim}[\frac{2\sin h/2\times \cos (x+h/2)}{h}\times \frac{1}{\sin x}]$

$=\underset{h\to 0}{lim}\frac{\log (1+\frac{\sin (x+h)-\sin x}{\sin x})}{\left(\frac{\sin (x+h)-\sin x}{\sin x}\right)}\times \underset{h\to 0}{lim}[\frac{\sin h/2\times \cos (x+h/2)}{h/2}\times \frac{1}{\sin x}]$

$=1\times \cos x\times \frac{1}{\sin x}$
$=\cot x$