Question

If f '(1) = 2 and g'$\left(\sqrt{2}\right)$ = 4, then the derivative of f(tan x) with respect of g(secx) at x = $\frac{\mathrm{\pi }}{4}$ is equal to ______________.

Answer 1

Let u(x) = f(tanx) and v(x) = g(secx).

$u\left(x\right)=f\left(\tan x\right)$

$\Rightarrow \frac{du}{dx}=\frac{d}{dx}f\left(\tan x\right)$

$\Rightarrow \frac{du}{dx}=f\text{'}\left(\tan x\right)\frac{d}{dx}\tan x$

$\Rightarrow \frac{du}{dx}=f\text{'}\left(\tan x\right)\times {\sec }^{2}x$ .....(1)

$v\left(x\right)=g\left(\sec x\right)$

$\Rightarrow \frac{dv}{dx}=\frac{d}{dx}g\left(\sec x\right)$

$\Rightarrow \frac{dv}{dx}=g\text{'}\left(\sec x\right)\frac{d}{dx}\sec x$

$\Rightarrow \frac{dv}{dx}=g\text{'}\left(\sec x\right)\times \sec x\tan x$ .....(2)

$\therefore \frac{du}{dv}=\frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}$

$\Rightarrow \frac{du}{dv}=\frac{{\sec }^{2}x\times f\text{'}\left(\tan x\right)}{\sec x\tan x\times g\text{'}\left(\sec x\right)}$

$\Rightarrow \frac{du}{dv}=\frac{\sec x\times f\text{'}\left(\tan x\right)}{\tan x\times g\text{'}\left(\sec x\right)}$

Putting $x=\frac{\mathrm{\pi }}{4}$, we get

${\left(\frac{du}{dv}\right)}_{x=\frac{\mathrm{\pi }}{4}}=\frac{\sec {\displaystyle \frac{\mathrm{\pi }}{4}}\times f\text{'}\left(\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\times g\text{'}\left(\sec {\displaystyle \frac{\mathrm{\pi }}{4}}\right)}$

$\Rightarrow {\left(\frac{du}{dv}\right)}_{x=\frac{\mathrm{\pi }}{4}}=\frac{\sqrt{2}\times f\text{'}\left(1\right)}{1\times g\text{'}\left(\sqrt{2}\right)}$

$\Rightarrow {\left(\frac{du}{dv}\right)}_{x=\frac{\mathrm{\pi }}{4}}=\frac{\sqrt{2}\times 2}{1\times 4}$

$\Rightarrow {\left(\frac{du}{dv}\right)}_{x=\frac{\mathrm{\pi }}{4}}=\frac{1}{\sqrt{2}}$

Thus, the derivative of f(tan x) with respect of g(secx) at x = $\frac{\mathrm{\pi }}{4}$ is $\frac{1}{\sqrt{2}}$.


If f '(1) = 2 and g'$\left(\sqrt{2}\right)$ = 4, then the derivative of f(tan x) with respect of g(secx) at x = $\frac{\mathrm{\pi }}{4}$ is equal to $\underline{\frac{1}{\sqrt{2}}}$.