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.

$f\left(x\right)={\sin }^2\left({\sec }^{1}x\right)$

Short answer

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

Step-by-step solution

Step 1. Given Information.

The given function is$f\left(x\right)={\sin }^2\left({\sec }^{-1}x\right)$.

Step 2. Preliminary Algebra.

We know,

$$\begin{gathered} ChainRule: \\ (f\circ u{)}^{\text{'}}(x)=f^{\text{'}}(u\left(x\right)\left)u^{\text{'}}\right(x) \\ Also, \\ \frac{d}{dx}{x}^n=n{x}^{n-1} \\ \frac{d}{dx}\left({\sec }^{-1}x\right)=\frac{1}{\left|x\right|\sqrt{x^2-1}} \\ \frac{d}{dx}(\sin x)=\cos x \end{gathered}$$

Step 3. Derivative of the function.

The derivative of the function is ,

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