Question

Cos^-1(1-9^x/1+9^x) differentiate with respect to x

Answer 1

SOLUTION

TO DIFFERENTIATE

$\displaystyle \sf{ { \cos}^{ - 1} \bigg( \: \frac{1 - {9}^{x} }{1 + {9}^{x} } \bigg) \: }$

EVALUATION

Let us take

$\displaystyle \sf{y = { \cos}^{ - 1} \bigg( \: \frac{1 - {9}^{x} }{1 + {9}^{x} } \bigg) \: }$

$\implies \displaystyle \sf{y = { \cos}^{ - 1} \bigg( \: \frac{1 - {( {3}^{x} )}^{2} }{1 + {({3}^{x} )}^{2} } \bigg) \: }$

Let $\sf{ {3}^{x} = \tan \theta}$

Then

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

$\implies \displaystyle \sf{y = { \cos}^{ - 1}( \cos2 \theta)\: }$

$\implies \displaystyle \sf{y = 2 \theta}$

$\implies \displaystyle \sf{y = 2 { { \tan}^{ - 1} ( {3}^{x} }) }$

Now Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{dy}{dx} =2. \frac{1}{1 + { ( {3}^{x} )}^{2} } \: . \: \frac{d}{dx} \bigg( {3}^{x} \bigg) }$

$\implies \displaystyle \sf{ \frac{dy}{dx} = \frac{2}{1 + {9}^{x} } \: . \: {3}^{x} \ln 3 }$

FINAL ANSWER

$\boxed{ \displaystyle \sf{ \: \frac{d}{dx } \bigg [\displaystyle \sf{ { \cos}^{ - 1} \bigg( \: \frac{1 - {9}^{x} }{1 + {9}^{x} } \bigg) \: } \bigg]= \frac{2}{1 + {9}^{x} } \: . \: {3}^{x} \ln 3 } \: }$

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