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Differentiate Sin − 1 { 2 X + 1 ⋅ 3 X 1 + ( 36 ) X } with Respect to X ? - CBSE (Arts) Class 12 - Mathematics

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Question

Differentiate \[\sin^{- 1} \left\{ \frac{2^{x + 1} \cdot 3^x}{1 + \left( 36 \right)^x} \right\}\]  with respect to x ?

Solution

We have ,

\[y = \sin^{- 1} \left\{ \frac{2^{x + 1} \cdot 3^x}{1 + \left( 36 \right)^x} \right\}\]

\[ \Rightarrow y = \sin^{- 1} \left\{ \frac{2 \cdot 6^x}{1 + 6^{2x}} \right\}\]

\[\text{ put } 6^x = \tan\theta\]

\[ \Rightarrow \theta = \tan^{- 1} \left( 6^x \right)\]

\[\text{ Now }, \]

\[y = \sin^{- 1} \left\{ \frac{2\tan\theta}{1 + \tan^2 \theta} \right\}\]

\[ \Rightarrow y = \sin^{- 1} \left\{ \sin2\theta \right\}\]

\[ \Rightarrow y = 2\theta\]

\[ \Rightarrow y = 2 \tan^{- 1} \left( 6^x \right)\]

\[ \Rightarrow \frac{dy}{dx} = 2 \times \frac{1}{\left( 6^x \right)^2} \times 6^x \log6\]

\[ \therefore \frac{dy}{dx} = \frac{2\left( \log6 \right) 6^x}{{36}^x}\]

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Solution Differentiate Sin − 1 { 2 X + 1 ⋅ 3 X 1 + ( 36 ) X } with Respect to X ? Concept: Simple Problems on Applications of Derivatives.
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