Question

Please find the derivative of $x3^{(2x^2-1)}$. Show all of your work and explain as much as possible.​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

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

$\boxed{ \bf \: \dfrac{d}{dx} k = 0}$

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

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

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

$\rm :\longmapsto\:Let \: y \: = \: x \: {3}^{ {2x}^{2} - 1}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}\bigg(x \: {3}^{ {2x}^{2} - 1} \bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = x\dfrac{d}{dx} {3}^{ {2x}^{2} - 1} + {3}^{ {2x}^{2} - 1} \dfrac{d}{dx}x$

$\rm :\longmapsto\:\dfrac{dy}{dx} = x{3}^{ {2x}^{2} - 1}log3\dfrac{d}{dx}( {2x}^{2} - 1) + {3}^{ {2x}^{2} - 1} \times 1$

$\rm :\longmapsto\:\dfrac{dy}{dx} = x{3}^{ {2x}^{2} - 1}log3(4x) + {3}^{ {2x}^{2} - 1}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {3}^{ {2x}^{2} - 1}(4 {x}^{2} log3+ 1)$

Additional Information :-

$\boxed{ \bf \: \dfrac{d}{dx} x = 1}$

$\boxed{ \bf \: \dfrac{d}{dx} sinx = cosx}$

$\boxed{ \bf \: \dfrac{d}{dx} cosx = - \: sinx}$

$\boxed{ \bf \: \dfrac{d}{dx} tanx = {sec}^{2} x}$

$\boxed{ \bf \: \dfrac{d}{dx} secxtanx = t anx}$

$\boxed{ \bf \: \dfrac{d}{dx} cotx = - {cosec}^{2} x}$

$\boxed{ \bf \: \dfrac{d}{dx} cosecx = - cotx \: cosecx}$

$\boxed{ \bf \: \dfrac{d}{dx} {e}^{x} = {e}^{x} }$

$\boxed{ \bf \: \dfrac{d}{dx} \dfrac{u}{v} = \dfrac{v\dfrac{d}{dx} u - u\dfrac{d}{dx}v }{ {v}^{2} }}$

Answer 2

Answer:

Lety=x3

2x

2

−1

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

\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}\bigg(x \: {3}^{ {2x}^{2} - 1} \bigg):⟼

dx

d

y=

dx

d

(x3

2x

2

−1

)

\rm :\longmapsto\:\dfrac{dy}{dx} = x\dfrac{d}{dx} {3}^{ {2x}^{2} - 1} + {3}^{ {2x}^{2} - 1} \dfrac{d}{dx}x:⟼

dx

dy

=x

dx

d

3

2x

2

−1

+3

2x

2

−1

dx

d

x

\rm :\longmapsto\:\dfrac{dy}{dx} = x{3}^{ {2x}^{2} - 1}log3\dfrac{d}{dx}( {2x}^{2} - 1) + {3}^{ {2x}^{2} - 1} \times 1:⟼

dx

dy

=x3

2x

2

−1

log3

dx

d

(2x

2

−1)+3

2x

2

−1

×1

\rm :\longmapsto\:\dfrac{dy}{dx} = x{3}^{ {2x}^{2} - 1}log3(4x) + {3}^{ {2x}^{2} - 1}:⟼

dx

dy

=x3

2x

2

−1

log3(4x)+3

2x

2

−1

\rm :\longmapsto\:\dfrac{dy}{dx} = {3}^{ {2x}^{2} - 1}(4 {x}^{2} log3+ 1):⟼

dx

dy

=3

2x

2

−1

(4x

2

log3+1)

Additional Information :-

\boxed{ \bf \: \dfrac{d}{dx} x = 1}

dx

d

x=1

\boxed{ \bf \: \dfrac{d}{dx} sinx = cosx}

dx

d

sinx=cosx

\boxed{ \bf \: \dfrac{d}{dx} cosx = - \: sinx}

dx

d

cosx=−sinx

\boxed{ \bf \: \dfrac{d}{dx} tanx = {sec}^{2} x}

dx

d

tanx=sec

2

x

\boxed{ \bf \: \dfrac{d}{dx} secxtanx = t anx}

dx

d

secxtanx=tanx

\boxed{ \bf \: \dfrac{d}{dx} cotx = - {cosec}^{2} x}

dx

d

cotx=−cosec

2

x

\boxed{ \bf \: \dfrac{d}{dx} cosecx = - cotx \: cosecx}

dx

d

cosecx=−cotxcosecx

\boxed{ \bf \: \dfrac{d}{dx} {e}^{x} = {e}^{x} }

dx

d

e

x

=e

x

\boxed{ \bf \: \dfrac{d}{dx} \dfrac{u}{v} = \dfrac{v\dfrac{d}{dx} u - u\dfrac{d}{dx}v }{ {v}^{2} }}

dx

d

v

u

=

v

2

v

dx

d

u−u

dx

d

v