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

ConceptSimple Problems on Applications of Derivatives

#### 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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