Question

Use implicit differentiation and the fact that $sec\left(se{c}^{-1}x\right)=x$for all $x$in the domain of $se{c}^{-1}x$to prove that $\frac{d}{dx}\left(se{c}^{-1}x\right)=\frac{1}{\left|x\right|\sqrt{x^2-1}}$. You will have to consider the cases$x>1$and$x<-1$ separately.

Short answer

We proved $\frac{d}{dx}\left(se{c}^{-1}x\right)=\frac{1}{\left|x\right|\sqrt{x^2-1}}$usingimplicit differentiation.

Step-by-step solution

Step 1. Given Information 

We are given $se{c}^{-1}x$ and using implicit differentiation we need to prove that $\frac{d}{dx}\left(se{c}^{-1}x\right)=\frac{1}{\left|x\right|\sqrt{x^2-1}}$.

Step 2. Proving the expression  

We consider $x>1$and $x<-1$,

$sec\left(se{c}^{-1}x\right)=x$

Differentiating both sides we get,

$$\begin{gathered} \frac{d}{dx}\left\{sec\left(se{c}^{-1}x\right)\right\}=\frac{d}{dx}\left(x\right) \\ \Rightarrow sec\left(se{c}^{-1}x\right)\tan \left(se{c}^{-1}x\right)\frac{d}{dx}\left(se{c}^{-1}x\right)=1 \\ \Rightarrow \frac{d}{dx}\left(se{c}^{-1}x\right)=\frac{1}{sec\left(se{c}^{-1}x\right)\tan \left(se{c}^{-1}x\right)} \\ \Rightarrow \frac{d}{dx}\left(se{c}^{-1}x\right)=\frac{1}{xt\mathrm{an}\left(se{c}^{-1}x\right)} \\ \Rightarrow \frac{d}{dx}\left(se{c}^{-1}x\right)=\frac{1}{x\left(\left|x\right|\sqrt{1-{\displaystyle \frac{1}{{\left(sec\left(se{c}^{-1}x\right)\right)}^2}}}\right)} \\ \Rightarrow \frac{d}{dx}\left(se{c}^{-1}x\right)=\frac{1}{x\left(\left|x\right|{\displaystyle \frac{\sqrt{x^2-1}}{x}}\right)} \\ \Rightarrow \frac{d}{dx}\left(se{c}^{-1}x\right)=\frac{1}{\left|x\right|\sqrt{x^2-1}} \end{gathered}$$