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

$f\left(x\right)=\ln \left(arcsec\left({\sin }^{2}x\right)\right)$

Short answer

The derivative is$\frac{2\sin x\cos x}{arcsec\left({\sin }^{2}x\right)}\left(\frac{1}{{\sin }^{2}x\sqrt{{\sin }^{4}x-1}}\right)$.

Step-by-step solution

Step 1. Given Information.

The given function is,

$f\left(x\right)=\ln \left(arcsec\left({\sin }^{2}x\right)\right)$.

Step 2. Preliminary Algebra.

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

Step 3. Derivative of the function.

The derivative of the function is,

$$\begin{gathered} \frac{d}{dx}\ln \left(arcsec\left({\sin }^{2}x\right)\right)=\frac{1}{arcsec\left({\sin }^{2}x\right)}\frac{d}{dx}\left(arcsec\left({\sin }^{2}x\right)\right) \\ =\frac{1}{arcsec\left({\sin }^{2}x\right)}\left(\frac{1}{\left|{\sin }^{2}x\right|\sqrt{{\left({\sin }^{2}x\right)}^2-1}}\right)\frac{d}{dx}\left({\sin }^{2}x\right) \\ =\frac{1}{arcsec\left({\sin }^{2}x\right)}\left(\frac{1}{\left|{\sin }^{2}x\right|\sqrt{{\sin }^{4}x-1}}\right)\left(2\sin x\right)\frac{d}{dx}\left(\sin x\right) \\ =\frac{1}{arcsec\left({\sin }^{2}x\right)}\left(\frac{1}{{\sin }^{2}x\sqrt{{\sin }^{4}x-1}}\right)\left(2\sin x\cos x\right) \end{gathered}$$

$=\frac{2\sin x\cos x}{arcsec\left({\sin }^{2}x\right)}\left(\frac{1}{{\sin }^{2}x\sqrt{{\sin }^{4}x-1}}\right)$