Question

How can the derivative of ${\sin }^{-1}x$be equal to both$\frac{1}{\sqrt{1-x^2}}\text{and}\frac{1}{\cos \left({\sin }^{-1}x\right)}$?Which expression is easier to use, and why?

Short answer

The expression$\frac{1}{\sqrt{1-x^2}}$is easier to use because it is entirely algebraic and derivative is easier to calculate at a number.

Step-by-step solution

Step 1. Given Information.

The given derivative of ${\sin }^{-1}x$is,

$\frac{1}{\sqrt{1-x^2}}\text{and}\frac{1}{\cos \left({\sin }^{-1}x\right)}$.

Step 2. Explanation.

Let,

$$\begin{gathered} {\sin }^{-1}x=\theta \\ \Rightarrow \sin \theta =x \\ Now, \\ \cos \theta =\sqrt{1-{\sin }^2\theta } \\ =\sqrt{1-x^2} \\ \theta ={\cos }^{-1}\sqrt{1-x^2} \\ {\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2} \end{gathered}$$

Also, we know,

$$\begin{gathered} \sin \left({\sin }^{-1}x\right)=x \\ \frac{d}{dx}\sin \left({\sin }^{-1}x\right)=\frac{dx}{dx} \\ \cos \left({\sin }^{-1}x\right)\frac{d}{dx}{\sin }^{-1}x=1 \\ \frac{d}{dx}{\sin }^{-1}x=\frac{1}{\cos \left({\cos }^{-1}\sqrt{1-x^2}\right)} \\ =\frac{1}{\sqrt{1-x^2}} \end{gathered}$$

Step 3. Easier derivative.

The derivative$\frac{1}{\sqrt{1-x^2}}$is easier to use because it is entirely algebraic and derivative is easier to calculate at a number.