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Solution for Differentiate Tan − 1 ( 2 X + 1 1 − 4 X ) , − ∞ < X < 0 ? - CBSE (Commerce) Class 12 - Mathematics

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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 for question: Differentiate Tan − 1 ( 2 X + 1 1 − 4 X ) , − ∞ < X < 0 ? concept: Simple Problems on Applications of Derivatives. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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