Question

The differential coefficient of the given function ${\log }_e\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$is

• A. $\cos ecx$
• B. $\sec x$
• C. $\tan x$
• D. $\cos x$

Answer 1

The correct option is B

$\sec x$

Explanation for the correct option:

Simplifying the function:

Simplifying the term inside log:

$\sqrt{\frac{1+\sin x}{1-\sin x}}=\sqrt{\frac{{\left(1+\sin x\right)}^2}{1-{\sin }^{2}x}}$ (multiplying and dividing by $1+\sin x$)

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

Therefore

$$\begin{gathered} {\log }_e\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)={\log }_e\left(\sqrt{\frac{{\left(1+\sin x\right)}^2}{{\cos }^{2}x}}\right) \\ ={\log }_e\left(\frac{1+\sin x}{\cos x}\right) \\ ={\log }_e\left(\sec x+\tan x\right) \end{gathered}$$

$$\begin{gathered} \frac{d}{dx}\left({\log }_e\left(\sec x+\tan x\right)\right)=\frac{\sec x\tan x+{\sec }^{2}x}{\sec x+\tan x} \\ =\sec x \end{gathered}$$ $$\begin{gathered} \left\{\left(\frac{d}{dx}\left(\log \left(f\left(x\right)\right)\right)\right)=\frac{1}{f\left(x\right)}f\text{'}\left(x\right) \\ \frac{d}{dx}\left(secx+\tan x\right)=secx.\tan x+se{c}^{2}x\right\} \end{gathered}$$

Hence, option (B) is the correct answer.