PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for Differentiate Tan − 1 ( 2 X + 1 1 − 4 X ) , − ∞ < X < 0 ? - PUC Karnataka Science Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Question

Differentiate $\tan^{- 1} \left( \frac{2^{x + 1}}{1 - 4^x} \right), - \infty < x < 0$ ?

#### Solution

$\text{ Let, y } = \tan^{- 1} \left\{ \frac{2^{x + 1}}{1 - 4^x} \right\}$

$\text{ put }2^x = \tan\theta$

$y = \tan^{- 1} \left\{ \frac{2^x \times 2}{1 - \left( 2^x \right)^2} \right\}$

$y = \tan^{- 1} \left( \frac{2 \tan\theta}{1 - \tan^2 \theta} \right)$

$y = \tan^{- 1} \left( \tan2\theta \right) . . . \left( i \right)$

$\text{ Here }, - \infty < x < 0$

$\Rightarrow 2^{- \infty} < 2^x < 2^\circ$

$\Rightarrow 0 < 2^x < 1$

$\Rightarrow 0 < \theta < \frac{\pi}{4}$

$\Rightarrow 0 < 2\theta < \frac{\pi}{2}$

$\text{ So, from equation } \left( i \right),$

$y = 2\theta \left[ \text{ Since }, \tan^{- 1} \left( \tan\theta \right) = \theta, if \theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right]$

$y = 2 \tan^{- 1} \left( 2^x \right)$

$\text{ Differentiating it with respect to x} ,$

$\frac{d y}{d x} = \frac{2}{1 + \left( 2^x \right)^2}\frac{d}{dx}\left( 2^x \right)$

$\frac{d y}{d x} = \frac{2 \times 2^x \log_e 2}{1 + 4^x}$

$\frac{d y}{d x} = \frac{2^{x + 1} \log_e 2}{1 + 4^x}$

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Solution Differentiate Tan − 1 ( 2 X + 1 1 − 4 X ) , − ∞ < X < 0 ? Concept: Simple Problems on Applications of Derivatives.
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