Question

Find $\frac{dy}{dx}=si{n}^{-1}x$

Answer 1

Let $y=f\left(x\right)={\sin }^{-1}x.$ so, $x=\sin y$

then,$y+k=f(x+h)={\sin }^{-1}(x+h)$ so, $x+h=\sin (y+k)$

again, as $h\to 0,k\to 0$

We know that, the first principal of derivative

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$={lim}_{h\to 0}\frac{{\sin }^{-1}(x+h)-{\sin }^{-1}\left(x\right)}{x+h-x}$

$={lim}_{k\to 0}\frac{(y+k)-\left(y\right)}{\sin (y+k)-\sin y}$

$={lim}_{k\to 0}\frac{k}{2\cos \frac{(y+k+y)}{2}\sin \frac{(y+k-y)}{2}}$

$={lim}_{k\to 0}\frac{k}{2\cos \frac{(2y+k)}{2}\sin \frac{k}{2}}$

$={lim}_{k\to 0}\frac{1}{2\cos \frac{(2y+k)}{2}}.{lim}_{k\to 0}\frac{k}{\sin \frac{k}{2}}$

$={lim}_{k\to 0}\frac{1}{2\cos \frac{(2y+k)}{2}}.{lim}_{k\to 0}\frac{\frac{2k}{2}}{\sin \frac{k}{2}}$

$={lim}_{k\to 0}\frac{2}{2\cos \frac{(2y+k)}{2}}.{lim}_{k\to 0}\frac{\frac{k}{2}}{\sin \frac{k}{2}}$

$={lim}_{k\to 0}\frac{2}{2\cos \frac{(2y+k)}{2}}.1$

Taking limit, and we get

$=\frac{2}{2\cos \frac{(2y+0)}{2}}.1$

$=\frac{1}{\cos \frac{2y}{2}}$

$=\frac{1}{\cos y}$

$=\frac{1}{\sqrt{1-{\sin }^{2}y}}$

$=\frac{1}{\sqrt{1-x^2}}$

Hence, this is the answer.