Question

Explain the differentiation of ${\sin }^{-1}x$.

Answer 1

Explain the differentiation of ${\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathit{x}$.

Let us assume that, $y={\sin }^{-1}x$.

Thus, to get the differentiation of ${\sin }^{-1}x$, we need to find $\frac{dy}{dx}$.

Take Sine of both sides of the equation.

$$\begin{gathered} \sin y=\sin \left({\sin }^{-1}x\right) \\ \Rightarrow \sin y=x \end{gathered}$$

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \frac{d\left(\sin y\right)}{dx}=\frac{dx}{dx} \\ \Rightarrow \cos y·\frac{dy}{dx}=1 \\ \Rightarrow \frac{dy}{dx}=\frac{1}{\cos y} \end{gathered}$$

It is known that, ${\sin }^{2}x+{\cos }^{2}x=1$.

As $\sin y=x$.

Thus, $\cos y=\sqrt{1-{\sin }^{2}y}$

$\Rightarrow \cos y=\sqrt{1-x^2}$

$\therefore \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$.

Therefore, the differentiation of ${\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathit{x}$ is $\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}$.