Question

differentiate wrt x (find dy/dx).​

Answer 1

SOLUTION

TO DETERMINE

To Differentiate with respect to x

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

EVALUATION

Here the given function is

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

$\displaystyle \sf{ \implies \: y = { \tan}^{ - 1} \bigg( \: \frac{1 + \sqrt{x} }{1 - \sqrt{x} } \bigg)}$

$\displaystyle \sf{ let \: \: x = { \tan}^{2} \theta }$

Then above becomes

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

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

$\displaystyle \sf{ \implies \: y = { \tan}^{ - 1} \bigg( \: \frac{ \tan \frac{\pi}{4} + { \tan}^{} \theta }{1 - \tan \frac{\pi}{4} { \tan}^{} \theta} \bigg)}$

$\displaystyle \sf{ \implies \: y = { \tan}^{ - 1} \tan \bigg( \frac{\pi}{4} + \theta \bigg) }$

$\displaystyle \sf{ \implies \: y = \bigg( \frac{\pi}{4} + \theta \bigg) }$

$\displaystyle \sf{Now \: \: \: x = { \tan}^{2} \: \theta \: \: implies \: \: \theta = { \tan}^{ - 1} \sqrt{x} }$

$\displaystyle \sf{ \therefore \: \: \: y = \bigg( \frac{\pi}{4} + { \tan}^{ - 1} \sqrt{x} \bigg) }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \: \frac{dy}{dx} = \frac{d}{dx} \bigg( \frac{\pi}{4} + { \tan}^{ - 1} \sqrt{x} \bigg) }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = \frac{d}{dx} \bigg( \frac{\pi}{4} \bigg) +\frac{d}{dx} \bigg( { \tan}^{ - 1} \sqrt{x} \bigg) }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = 0 + \frac{1}{1 + {( \sqrt{x} )}^{2} } . \frac{d}{dx}( \sqrt{x} )}$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = \frac{1}{2 \sqrt{x} (1 + x)} }$

FINAL ANSWER

$\boxed{ \: \: \displaystyle \sf{ \: \frac{dy}{dx} = \frac{1}{2 \sqrt{x} (1 + x)} } \: \: }$

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