Question

Find the derivative of $\sin x$ at $x=0.$

Answer 1

Let $f\left(x\right)=\sin x$
We know that $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\sin (x+h)-\sin x}{h}$
Putting $x=0$
$\Rightarrow f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{\sin (0+h)-\sin \left(0\right)}{h}$
$\Rightarrow f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{\sin h-0}{h}$
$\Rightarrow f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{\sin h}{h}$
$\Rightarrow f^{\prime }\left(0\right)=1$
Hence, derivative of $\sin x$ at $x=0$ is $1$.