Question

differentiate tan^-1 [√1+a^2x^2 -1÷ax]​

Answer 1

Answer:

given function is

$y = {tan}^{ - 1} ( \frac{ \sqrt{1 + {a}^{2} {x}^{2} } - 1 }{ax} )$

☑ let us put ax = tan theta, then theta = tan^(-1)ax

therefor,

$y = {tan}^{ - 1} ( \frac{ \sqrt{1 + {tan}^{2} theta} - 1}{tantheta} )$

$= {tan}^{ - 1} ( \frac{ \sqrt{ {sec}^{2}theta } - 1}{tan \: theta} )$

$= {tan}^{ - 1} ( \frac{sec \: theta - 1}{tan \: theta} )$

$= {tan}^{ - 1} ( \frac{ \frac{1}{cos \: theta} - 1}{ \frac{sin \: theta}{cos \: theta} } )$

$= {tan}^{ - 1} ( \frac{1 - cos \: theta}{sin \: theta} )$

=${tan}^{ - 1} ( \frac{2 {sin}^{2} \frac{theta}{2} }{2sin \frac{theta}{2} \times cos \frac{theta}{2} } )$

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

$= \frac{theta}{2}$

☑ Now put the value of theta ,Therefor the equation becomes,

$y = \frac{1}{2} {tan}^{ - 1} (ax)$

☑ Differentiate wrt x

$\frac{dy}{dx} = \frac{d}{dx} ( \frac{1}{2} {tan}^{ - 1} (ax))$

$= \frac{1}{2} \frac{d}{dx} ( {tan}^{ - 1} (ax))$

$= \frac{1}{2} \times \frac{1}{1 + ({ax})^{2} } \times \frac{d}{dx} (ax)$

$= \frac{1}{2(1 + {a}^{2} {x}^{2}) } \times a$

Therefor,

$\frac{dy}{dx} = \frac{1}{2} ( \frac{a}{1 + {a}^{2} {x}^{2} } )$

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Answer 2

Step-by-step explanation:

step by step explanations