Question

The derivative of $f(\tan x)$ with respect to $g(secx)$ at $x=\frac{\pi }{4}$, where $f\text{'}\left(1\right)=2$ and $g\text{'}\left(\sqrt{2}\right)=4$, is

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $1$
• D. None of these

Answer 1

The correct option is A

$\frac{1}{\sqrt{2}}$

Explanation for the correct option:

Finding the derivative of $f(\tan x)$ with respect to $g(secx)$:

Let

$$\begin{gathered} \upsilon =f\left(\tan x\right),v=g\left(secx\right) \\ \frac{d}{dx}\left(u\right)=\frac{du}{dx} \\ =\frac{d}{dx}\left(f\left(\tan x\right)\right) \\ =\frac{d\left(f\left(\tan x\right)\right)}{d\left(\tan x\right)}x\frac{d\left(\tan x\right)}{dx}\left(Chain\mathit{}Rule\right) \\ =f\text{'}\left(\tan x\right)x\left(se{c}^{2}x\right) \\ =f\text{'}\left(\tan x\right)\left(se{c}^{2}x\right) \end{gathered}$$

Now,

$$\begin{gathered} \frac{d}{dx}\left(v\right)=\frac{dv}{dx} \\ =\frac{d}{dx}\left(g\left(secx\right)\right) \\ =\frac{d\left(g\left(secx\right)\right)}{d\left(secx\right)}x\frac{d\left(secx\right)}{dx}\left(Chain\mathit{}Rule\right) \\ =g\text{'}\left(secx\right)xsecx\tan x \end{gathered}$$

Now,

$\frac{du}{dv}=\left(\frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}\right)=\frac{f\text{'}\left(\tan x\right)se{c}^{2}x}{g\text{'}\left(secx\right)secx\tan x}=\frac{f\text{'}\left(\tan x\right)secx}{g\text{'}\left(secx\right)\tan x}$

$$\begin{gathered} \mathrm{Given}\mathrm{that}x=\frac{\mathrm{\pi }}{4} \\ \frac{du}{dv}=\frac{f\text{'}\left(\tan {\displaystyle \frac{\pi }{4}}\right)sec{\displaystyle \frac{\pi }{4}}}{g\text{'}sec\frac{\pi }{4}\tan \frac{\pi }{4}} \\ =\frac{f\text{'}\left(1\right)\sqrt{2}}{g\text{'}\left(\sqrt{2}\right)1} \\ =\frac{2\sqrt{2}}{\left(4\right)1} \\ =\frac{\sqrt{2}}{2} \\ =\frac{\sqrt{2}}{\sqrt{2\times \sqrt{2}}} \\ =\frac{1}{\sqrt{2}} \end{gathered}$$

Therefore, the correct answer is option (A).