Question

y=x³ sin^-1x; find dy/dx​

Answer 1

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

Given that

$\rm :\longmapsto\:y = {x}^{3} \: {sin}^{ - 1}x$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}( {x}^{3} \: {sin}^{ - 1}x)$

We know,

Product rule of differentiation

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

Now,

Here,

$\rm :\longmapsto\:u = {x}^{3}$

and

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

On substituting the values, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {sin}^{ - 1}x \: \dfrac{d}{dx} {x}^{3} \: + \: {x}^{3} \:\dfrac{d}{dx} {sin}^{ - 1}x$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {sin}^{ - 1}x \: (3 {x}^{3 - 1}) \: + \: {x}^{3} \:\dfrac{1}{ \sqrt{1 - {x}^{2} } }$

$\red{\bigg \{\tt \: \because\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: and \: \: \dfrac{d}{dx} {sin}^{ - 1}x = \dfrac{1}{ \sqrt{1 - {x}^{2} } } \bigg \}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 3 \: {sin}^{ - 1}x \: ({x}^{2}) \: + \:\dfrac{ {x}^{3} }{ \sqrt{1 - {x}^{2} } }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {x}^{2} \bigg(3 \: {sin}^{ - 1}x \: + \:\dfrac{ {x} }{ \sqrt{1 - {x}^{2} } } \bigg)$

Additional information :-

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}$

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

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

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

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

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

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

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

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

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

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

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