Question

Differentiate: $se{c}^{-1}\left(x\right)$

Answer 1

Differentiating $se{c}^{-1}\left(x\right)$:

Let us predict that $y=se{c}^{-1}\left(x\right)$

By rewriting in terms of secant, we get

$secy=x$

On differentiating with respect to $x$

$$\begin{gathered} \Rightarrow \frac{d}{dx}secy=\frac{dx}{dx} \\ \Rightarrow \frac{d}{dx}secy=1 \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{d}{dx}secy\times \frac{dy}{dy}=1 \\ \Rightarrow \frac{d}{dy}secy\times \frac{dy}{dx}=1 \\ \Rightarrow secy\times \tan y\times \frac{dy}{dx}=1 \end{gathered}$$

$\Rightarrow \frac{dy}{dx}=\frac{1}{secy\times \tan y}$

But as we know, $\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }\mathbf{=}\sqrt{\mathbf{s}\mathbf{e}{\mathbf{c}}^{\mathbf{2}}\mathbf{\theta }\mathbf{-}\mathbf{1}}$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{1}{(\sqrt{se{c}^2}y-1)\times secy} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{\left(\sqrt{x^2-1}\right)\times \times secy} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{\left(\sqrt{x^2-1}\right)\times x} \end{gathered}$$

Hence, differentiation of $se{c}^{-1}\left(x\right)$ is $\frac{\mathbf{1}}{\mathbf{\left(}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\mathbf{\right)}\mathbf{\times }\mathbf{x}}$ .