Chapter 3.6, Problem 19E

To determine

To calculate: The derivative of the function $y={\sin }^{-1}\left(2x+1\right)$ .

Answer to Problem 19E

The derivative of the function is $\frac{1}{\sqrt{-x^2-x}}$ .

Explanation of Solution

Given information:

The function $y={\sin }^{-1}\left(2x+1\right)$ .

Formula used:

Thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Differentiation rule for inverse trigonometric functions is $\frac{d}{dx}\left({\sin }^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}$ .

Calculation:

Consider the function $y={\sin }^{-1}\left(2x+1\right)$ .

Differentiate both sides with respect to x ,

  $\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left({\sin }^{-1}\left(2x+1\right)\right)$

Recall that differentiation rule for inverse trigonometric functions is $\frac{d}{dx}\left({\sin }^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}$ and chain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

  $\begin{array}{c}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left({\sin }^{-1}\left(2x+1\right)\right)\\ y\text{'}=\frac{1}{\sqrt{1-{\left(2x+1\right)}^2}}\cdot \left(2\right)\\ y\text{'}=\frac{2}{\sqrt{1-4x^2-4x-1}}\\ y\text{'}=\frac{2}{\sqrt{4\left(-x^2-x\right)}}\end{array}$

Simplify it further as,

  $\begin{array}{c}y\text{'}=\frac{2}{\sqrt{4\left(-x^2-x\right)}}\\ y\text{'}=\frac{2}{2\sqrt{-x^2-x}}\\ y\text{'}=\frac{1}{\sqrt{-x^2-x}}\end{array}$

Thus, the derivative of the function is $\frac{1}{\sqrt{-x^2-x}}$ .