Question

Find the derivatives of each of the functions in Exercises 17–50. In some cases it may be convenient to do some preliminary algebra

$f\left(x\right)=\frac{{\sin }^{-1}x}{{\sec }^{-1}x}$

Short answer

The derivative is$\frac{\frac{{\sec }^{-1}x}{\sqrt{1-x^2}}-\frac{{\sin }^{-1}x}{\left|x\right|\sqrt{x^2-1}}}{{\left({\sec }^{-1}x\right)}^2}$.

Step-by-step solution

Step 1. Given Information.

The given function is$f\left(x\right)=\frac{{\sin }^{-1}x}{{\sec }^{-1}x}$.

Step 2. Preliminary Algebra.

We know,

$$\begin{gathered} {\left(\frac{f}{g}\right)}^{\text{'}}\left(x\right)=\frac{f^{\text{'}}\left(x\right)g\left(x\right)-f\left(x\right)g^{\text{'}}\left(x\right)}{\left(g\right(x){)}^2} \\ \frac{d}{dx}\left({\sin }^{-1}x\right)=\frac{1}{\sqrt{1-x^2}} \\ \frac{d}{dx}\left({\sec }^{-1}x\right)=\frac{1}{\left|x\right|\sqrt{x^2-1}} \end{gathered}$$

Step 3. Derivative of the function.

The derivative of the function will be,

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