Question

Compute the derivative of $\sin x$.

Answer 1

Let $f\left(x\right)=\sin x$
We know that $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\sin (x+h)-\sin x}{h}$

Using $\{\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\cdot \sin \left(\frac{A-B}{2}\right)\}$
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{2\cos \left(\frac{x+h+x}{2}\right)\cdot \sin \left(\frac{x+h-x}{2}\right)}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}(\cos \left(\frac{2x+h}{2}\right)\cdot \frac{\sin \frac{h}{2}}{\frac{h}{2}})$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\cos \frac{2x+h}{2}\cdot \underset{h\to 0}{lim}\frac{\sin \frac{h}{2}}{\frac{h}{2}}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\cos \frac{2x+h}{2}\times 1$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\cos \frac{2x+h}{2}$
$\Rightarrow f^{\prime }\left(x\right)=\cos \left(\frac{2x+0}{2}\right)$
$\Rightarrow f^{\prime }\left(x\right)=\cos \left(\frac{2x}{2}\right)$
$\Rightarrow f^{\prime }\left(x\right)=\cos x$
$\therefore f^{\prime }\left(x\right)=\cos x$