Question

Differentiate the tan^-1[(√x+√a)/(1-√ax)] with respect to x.

Answer 1

Answer:

$\boxed{\boxed{\green{\huge \bold{\dfrac{1}{2 \sqrt{x} \: ( \: 1 + x \: )} \: }} }}$

Step-by-step explanation:

$\bold{Given} = > {tan}^{ - 1} ( \dfrac{ \sqrt{x} \: + \: \sqrt{a} }{1 \: - \: \sqrt{ax} } \: )$

$\bold{To \: find }\: = > derivative \: of \: given \: function$

$\bold{Concept \: used} = > \\ \green{ 1) \tan( \alpha + \beta ) \: = \dfrac{tan \alpha + tan \beta }{1 - tan \alpha \: tan \beta }}$

$\red{2) \dfrac{d}{dx} ( {tan}^{ - 1} x) \: = \dfrac{1}{1 + {x}^{2} }}$

$\blue{3)\dfrac{d}{ dx } ( \sqrt{x} \: ) = \dfrac{1}{2 \sqrt{x} }}$

$\bold{Solution} = > \\ let \\ y = {tan}^{ - 1} ( \: \dfrac{ \sqrt{x} + \sqrt{a} }{1 - \sqrt{ax} } \: )$

$= > y = {tan}^{ - 1} ( \: \dfrac{ \sqrt{x} + \sqrt{a} }{1 - \sqrt{x} \sqrt{a} } \: )$

$let \: \: tan \alpha = \sqrt{x} = > \alpha = {tan}^{ - 1} ( \sqrt{x} \: ) \\ \: \: \: \: \: \: \: tan \beta = \sqrt{a} = > \beta = {tan}^{ - 1} ( \sqrt{a} \: )$

$= > y = {tan}^{ - 1} ( \: \dfrac{tan \alpha + tan \beta }{1 - tan \alpha \: tan \beta } \: )$

$= > y = {tan}^{ - 1} tan( \alpha + \beta )$

$= > y = \alpha \: + \: \beta$

$= > y \: = {tan}^{ - 1} ( \: \sqrt{x} \: ) \: + {tan}^{ - 1} ( \: \sqrt{a} \: )$

$differentiating \: with \: respect \: to \: x$

$= > \dfrac{dy}{dx} = \dfrac{d}{dx} ( \: {tan}^{ - 1} \sqrt{x} \: ) \: + \dfrac{d}{dx} \: ( \: {tan}^{ - 1} \sqrt{a} \: )$

$=>\dfrac{dy}{dx} = \dfrac{1}{1 + ( \: \sqrt{x} \: ) ^{2} } \dfrac{d}{dx} ( \: \sqrt{x} \: ) \: + \: \: 0$

$= > \dfrac{dy}{dx} \: = \dfrac{1}{1 + x} ( \: \dfrac{1}{2 \sqrt{x} } \: )$

$= > \dfrac{dy}{dx} \: = \dfrac{1}{2 \sqrt{x} \: ( \: 1 \: + \: x \: )}$