Question

Differentiate the following w.r.t $x$ :

$\tan^{- 1} \{\frac{e^{2 x} + 1}{e^{2 x} - 1}\}$

Accepted Answer

We have, $\tan^{- 1} \{\frac{e^{2 x} + 1}{e^{2 x} - 1}\}$

$= \tan^{- 1} \{\frac{1 + e^{2 x}}{- (1 - e^{2 x})}\}$ $= - \tan^{- 1} \{\frac{1 + e^{2 x}}{1 - e^{2 x}}\}$

Putting $e^{2 x} = \tan θ$

$θ = \tan^{- 1} (e^{2 x}) . . . . . . . . . (i)$

$\Rightarrow - \tan^{- 1} \{\frac{1 + e^{2 x}}{1 - e^{2 x}}\} = - \tan^{- 1} \{\frac{1 + \tan θ}{1 - \tan θ}\}$

$= - \tan^{- 1} \{\frac{\tan \frac{π}{4} + \tan θ}{1 - \tan \frac{π}{4} \tan θ}\}$

$= - \tan^{- 1} \{\tan (\frac{π}{4} + θ)\}$

$= - (\frac{π}{4} + θ)$ $= - \frac{π}{4} - θ$

$\Rightarrow - \frac{π}{4} - \tan^{- 1} (e^{2 x})$ [From $(i)$ ]

Now, $\tan^{- 1} \{\frac{e^{2 x} + 1}{e^{2 x} - 1}\} = - \frac{π}{4} - \tan^{- 1} (e^{2 x})$

Now Differentiating, we get

$\frac{d (- \frac{π}{4} - \tan^{- 1} (e^{2 x}))}{d x}$

$= \frac{d (- \frac{π}{4})}{d x} - \frac{d (\tan^{- 1} (e^{2 x}))}{d x}$

$= 0 - \frac{d (\tan^{- 1} (e^{2 x}))}{d e^{2 x}} \frac{d e^{2 x}}{d 2 x} \frac{d 2 x}{d x}$

$= - (\frac{1}{1 + {(e^{2 x})}^{2}}) e^{2 x} \times 2$

$= \frac{- 2 e^{2 x}}{1 + e^{4 x}}$

Hence, $\frac{d}{d x} (\tan^{- 1} \{\frac{e^{2 x} + 1}{e^{2 x} - 1}\}) = \frac{- 2 e^{2 x}}{1 + e^{4 x}}$

R S Aggarwal Solutions for Exercise 4: EXERCISE

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 4: EXERCISE

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