Question

$please \: find \: the \: derivative \: of \: the \: above \: picture$
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Answer 1

Answer:

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Answer 2

Solution 1:-

Given,

$\longrightarrow y=\dfrac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}$

Dividing numerator and denominator by $e^{2x},$

$\longrightarrow y=\dfrac{1+e^{-4x}}{1-e^{-4x}}\quad\quad\dots(1)$

Substitute,

$\longrightarrow \tan\theta=e^{-4x}$

$\longrightarrow \theta=\tan^{-1}\left(e^{-4x}\right)$

$\longrightarrow\dfrac{d\theta}{dx}=\dfrac{-4e^{-4x}}{1+(e^{-4x})^2}$

$\longrightarrow\dfrac{d\theta}{dx}=\dfrac{-4e^{-4x}}{1+e^{-8x}}$

Then (1) becomes,

$\longrightarrow y=\dfrac{1+\tan\theta}{1-\tan\theta}$

$\longrightarrow y=\tan\left(\theta+\dfrac{\pi}{4}\right)\quad\quad\dots(2)$

From (1) and (2),

$\longrightarrow\tan\left(\theta+\dfrac{\pi}{4}\right)=\dfrac{1+e^{-4x}}{1-e^{-4x}}\quad\quad\dots(3)$

Then the derivative of y is,

$\longrightarrow\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\tan\left(\theta+\dfrac{\pi}{4}\right)\right)$

$\longrightarrow\dfrac{dy}{dx}=\dfrac{d}{d\theta}\left(\tan\left(\theta+\dfrac{\pi}{4}\right)\right)\cdot\dfrac{d\theta}{dx}$

$\longrightarrow\dfrac{dy}{dx}=\sec^2\left(\theta+\dfrac{\pi}{4}\right)\cdot\dfrac{-4e^{-4x}}{1+e^{-8x}}$

$\longrightarrow\dfrac{dy}{dx}=\left(1+\tan^2\left(\theta+\dfrac{\pi}{4}\right)\right)\cdot\dfrac{-4e^{-4x}}{1+e^{-8x}}$

From (3),

$\longrightarrow\dfrac{dy}{dx}=\left(1+\left(\dfrac{1+e^{-4x}}{1-e^{-4x}}\right)^2\right)\cdot\dfrac{-4e^{-4x}}{1+e^{-8x}}$

$\longrightarrow\dfrac{dy}{dx}=\dfrac{(1-e^{-4x})^2+(1+e^{-4x})^2}{(1-e^{-4x})^2}\cdot\dfrac{-4e^{-4x}}{1+e^{-8x}}$

$\longrightarrow\dfrac{dy}{dx}=\dfrac{2(1+e^{-8x})}{(1-e^{-4x})^2}\cdot\dfrac{-4e^{-4x}}{1+e^{-8x}}\quad\left[\because(a-b)^2+(a+b)^2=2(a^2+b^2)\right]$

$\longrightarrow\dfrac{dy}{dx}=-\dfrac{8e^{-4x}}{(1-e^{-4x})^2}$

Multiplying numerator and denominator by $e^{4x},$

$\longrightarrow\dfrac{dy}{dx}=-\dfrac{8e^{-4x}\cdot e^{4x}}{(1-e^{-4x})^2\cdot e^{4x}}$

$\longrightarrow\dfrac{dy}{dx}=-\dfrac{8}{\left((1-e^{-4x})\cdot e^{2x}\right)^2}$

$\longrightarrow\underline{\underline{\dfrac{dy}{dx}=-\dfrac{8}{\left(e^{2x}-e^{-2x}\right)^2}}}}}$

Solution 2:-

Given,

$\longrightarrow y=\dfrac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\quad\quad\dots(4)$

Since $\coth x=\dfrac{e^x+e^{-x}}{e^x-e^{-x}},$

$\longrightarrow y=\coth(2x)\quad\quad\dots(5)$

From (4) and (5),

$\longrightarrow\dfrac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}=\coth(2x)\quad\quad\dots(6)$

Then the derivative of y is,

$\longrightarrow\dfrac{dy}{dx}=\dfrac{d}{dx}[\coth(2x)]$

$\longrightarrow\dfrac{dy}{dx}=\dfrac{d}{d(2x)}[\coth(2x)]\cdot\dfrac{d(2x)}{dx}$

$\longrightarrow\dfrac{dy}{dx}=\left(1-\coth^2(2x)\right)\cdot2$

$\longrightarrow\dfrac{dy}{dx}=2\left(1-\coth(2x)\right)\left(1+\coth(2x)\right)$

From (6),

$\longrightarrow\dfrac{dy}{dx}=2\left(1-\dfrac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\right)\left(1+\dfrac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\right)$

$\longrightarrow\dfrac{dy}{dx}=2\cdot\dfrac{-2e^{-2x}}{e^{2x}-e^{-2x}}\cdot\dfrac{2e^{2x}}{e^{2x}-e^{-2x}}$

$\longrightarrow\underline{\underline{\dfrac{dy}{dx}=-\dfrac{8}{\left(e^{2x}-e^{-2x}\right)^2}}}$