Question

if f(x)=x tan^-1(x) then f'(√3)= ?​

Answer 1

f(x)=xtan^-1(x). (given)

apply product rule,

f'(x)=x/1+x^2+tan^-1(x)

f'(√3)=√3/4+π/3

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:f(x) = x \: {tan}^{ - 1}x$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx}(x \: {tan}^{ - 1}x )$

We know,

Product Rule of Differentiation

$\rm :\longmapsto\:\dfrac{d}{dx}(u.v) = \: u \: \dfrac{d}{dx}v \: + \: v \: \dfrac{d}{dx}u$

So,

Here,

$\rm :\longmapsto\:u \: = \: x$

$\rm :\longmapsto\:v \: = \: {tan}^{ - 1}x$

On substituting the values in above formula, we get

$\rm :\longmapsto\:f'(x) = x \: \dfrac{d}{dx} {tan}^{ - 1}x \: + \: {tan}^{ - 1}x \: \dfrac{d}{dx}x$

$\rm :\longmapsto\:f'(x) = x \times \dfrac{1}{1 + {x}^{2} } \: + \: {tan}^{ - 1}x \times 1$

$\red{\bigg \{ \because \: \tt \: \dfrac{d}{dx}x = 1\bigg \}} \\ \red{\bigg \{ \because \: \tt \: \dfrac{d}{dx} {tan}^{ - 1}x = \dfrac{1}{1 + {x}^{2} }\bigg \}}$

$\rm :\longmapsto\:f'(x) = \dfrac{x}{1 + {x}^{2} } \: + \: {tan}^{ - 1}x$

$\rm :\longmapsto\:f'( \sqrt{3} ) = \dfrac{\sqrt{3} }{1 + {( \sqrt{3} )}^{2} } \: + \: {tan}^{ - 1}( \sqrt{3} )$

$\rm :\longmapsto\:f'( \sqrt{3} ) = \dfrac{\sqrt{3} }{1 +3} \: + \: {tan}^{ - 1}\bigg(tan\dfrac{\pi}{3} \bigg)$

$\rm :\longmapsto\:f'( \sqrt{3} ) = \dfrac{\sqrt{3} }{4} \: + \: \dfrac{\pi}{3}$

Additional information :-

$\red{\rm :\longmapsto\:\dfrac{d}{dx}x = 1}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}k = 0}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} } }$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}sinx = cosx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cosx = - \: sinx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cosecx = - \: cosecx \: cotx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}secx = \: secx \: tanx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}tanx = \: sec^{2} x}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cotx = - \: cosec^{2} x}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {sin}^{ - 1}x = \dfrac{1}{ \sqrt{1 - {x}^{2} } } }$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {cos}^{ - 1}x = \dfrac{ - \: 1}{ \sqrt{1 - {x}^{2} } } }$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1}x = \dfrac{ - \: 1}{ x \: \sqrt{{x}^{2} - 1} } }$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {sec}^{ - 1}x = \dfrac{\: 1}{ x \: \sqrt{{x}^{2} - 1} } }$