Question

Differentiate sin^-1 (2^x+1/1+4^x)
with respect to x
​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

We are aware of the Trigonometric formula that

FORMULA TO BE IMPLEMENTED

$\displaystyle \: \sin 2 \theta \: = \frac{2 \tan \theta}{1 + { \tan}^{2} \theta}$

TO DIFFERENTIATE

$\displaystyle \: { \sin}^{ - 1} \bigg( \frac{ {2}^{x + 1} }{1 + {4}^{x} } \bigg)$

EVALUATION

Let

$y = \displaystyle \: { \sin}^{ - 1} \bigg( \frac{ {2}^{x + 1} }{1 + {4}^{x} } \bigg)$

$= \displaystyle \: { \sin}^{ - 1} \bigg( \frac{ 2 \times {2}^{x} }{1 + {( {2}^{x} )}^{2} } \bigg) \: \: ...(1)$

Let

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

So that

$\theta = { \tan}^{ - 1} ( {2}^{x} )$

Therefore From Equation (1)

$y = \displaystyle \: { \sin}^{ - 1} \bigg( \frac{2 \tan \theta}{1 + { \tan}^{2} \theta} \bigg)$

$\implies \: y = { \sin}^{ - 1} ( \sin 2 \theta)$

$\implies \: y = 2 \theta$

$\implies \: y = 2 { \tan}^{ - 1} ( {2}^{x} )$

Differentiating both sides with respect to x

$\displaystyle \: \frac{dy}{dx} = 2 \times \frac{ {2}^{x} \ln2}{1 + {( {2}^{x}) }^{2} }$

$\implies \: \displaystyle \: \frac{dy}{dx} = \frac{ {2}^{x + 1} \ln2}{1 + {4}^{x} }$

RESULT

The required answer is

$\boxed{ \sf{ \: \: \displaystyle \: \frac{ {2}^{x + 1} \ln2}{1 + {4}^{x} } \: \: \: }}$