Question

a^x^2.b^x^3 derivative of the function​

Answer 1

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

The given function is

$\rm :\longmapsto\: {a}^{ {x}^{2} } \: {b}^{ {x}^{3} }$

Let assume that,

$\rm :\longmapsto\:y = {a}^{ {x}^{2} } \: {b}^{ {x}^{3} }$

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

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx}( {a}^{ {x}^{2} } \: {b}^{ {x}^{3} })$

We know,

$\underbrace{ \boxed{ \bf \: \dfrac{d}{dx}u.v = u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u}}$

So, using this we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = { {a}^{x} }^{2} \dfrac{d}{dx} { {b}^{x} }^{3} + {b}^{ {x}^{3} }\dfrac{d}{dx} {a}^{ {x}^{2} }$

We know,

$\underbrace{ \boxed{ \bf \: \dfrac{d}{dx} {a}^{x} = {a}^{x}loga}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = { {a}^{x} }^{2}{ {b}^{x} }^{3}logb \dfrac{d}{dx} {x}^{3} + {b}^{ {x}^{3} }{ {a}^{x} }^{2}loga\dfrac{d}{dx} {x}^{2}$

We know,

$\underbrace{ \boxed{ \bf \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = { {a}^{x} }^{2}{ {b}^{x} }^{3}logb({3x}^{2}) + {b}^{ {x}^{3} }{ {a}^{x} }^{2}loga(2x)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = { {a}^{x} }^{2}{ {b}^{x} }^{3}\bigg( {3x}^{2}logb \: + \: 2xloga \bigg)$

Alternative Method :-

$\rm :\longmapsto\:y = {a}^{ {x}^{2} } \: {b}^{ {x}^{3} }$

Taking log on both sides, we get

$\rm :\longmapsto\:logy = log( {a}^{{x}^{2}} \:{b}^{ {x}^{3}})$

We know,

$\underbrace{ \boxed{ \bf \: log(xy) = logx + logy}}$

So using this, we get

$\rm :\longmapsto\:logy = log{ {a}^{x} }^{2} + log{ {b}^{x} }^{3}$

We know,

$\underbrace{ \boxed{ \bf \: log( {x}^{y} ) = y log(x)}}$

Using this, we get

$\rm :\longmapsto\:logy = {x}^{2}loga + {x}^{3}logb$

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

$\rm :\longmapsto\:\dfrac{d}{dx}logy =\dfrac{d}{dx} {x}^{2}loga + \dfrac{d}{dx} {x}^{3}logb$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{d}{dx}y =loga\dfrac{d}{dx} {x}^{2}+logb \dfrac{d}{dx} {x}^{3}$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} =loga(2x) + logb( {3x}^{2} )$

$\rm :\longmapsto\:\dfrac{dy}{dx} = y(2xloga + {3x}^{2}logb)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = { {a}^{x} }^{2}{ {b}^{x} }^{3}\bigg( {3x}^{2}logb \: + \: 2xloga \bigg)$