Question

If y = cos^-1 ( 1 - 4^x / 1 + 4^x ) . Find dy/dx.

Answer 1

Given,

$\longrightarrow y=\cos^{-1}\left(\dfrac{1-4^x}{1+4^x}\right)$

$\longrightarrow y=\cos^{-1}\left(\dfrac{1-(2^2)^x}{1+(2^2)^x}\right)$

$\longrightarrow y=\cos^{-1}\left(\dfrac{1-(2^x)^2}{1+(2^x)^2}\right)\quad\quad\dots(1)$

Substitute,

$\longrightarrow 2^x=\tan\theta$

$\longrightarrow 2^x\log 2\ dx=\sec^2\theta\ d\theta$

$\longrightarrow \dfrac{d\theta}{dx}=\dfrac{2^x\log2}{\sec^2\theta}$

$\longrightarrow \dfrac{d\theta}{dx}=\dfrac{2^x\log2}{1+\tan^2\theta}$

$\longrightarrow \dfrac{d\theta}{dx}=\dfrac{2^x\log2}{1+(2^x)^2}$

$\longrightarrow \dfrac{d\theta}{dx}=\dfrac{2^x\log2}{1+4^x}\quad\quad\dots(2)$

Then (1) becomes,

$\longrightarrow y=\cos^{-1}\left(\dfrac{1-\tan^2\theta}{1+\tan^2\theta}\right)$

$\longrightarrow y=\cos^{-1}\left(\cos(2\theta)\right)$

$\longrightarrow y=2\theta$

$\longrightarrow dy=2\ d\theta$

Dividing by $dx,$

$\longrightarrow \dfrac{dy}{dx}=\dfrac{2\ d\theta}{dx}$

From (2),

$\longrightarrow \dfrac{dy}{dx}=2\times\dfrac{2^x\log2}{1+4^x}$

$\longrightarrow\underline{\underline{\dfrac{dy}{dx}=\dfrac{2^{x+1}\log2}{1+4^x}}}$