Question

Find third differential coefficient of x^4.e^2x​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\sf{y=x^4\,e^{2x}}$

$\sf{\implies\,\dfrac{dy}{dx}=4x^3\,e^{2x}+2x^4\,e^{2x}}$

$\sf{\implies\,\dfrac{dy}{dx}=2x^4\,e^{2x}\left(\dfrac{2}{x}+1\right)}$

$\sf{\implies\,\dfrac{dy}{dx}=2y\left(1+\dfrac{2}{x}\right)}$

$\sf{\implies\,\dfrac{d^2y}{dx^2}=2\left(1+\dfrac{2}{x}\right)\dfrac{dy}{dx}+2y\left(-\dfrac{2}{x^2}\right)}$

$\sf{\implies\,\dfrac{d^2y}{dx^2}=2y\left(1+\dfrac{2}{x}\right)^2-y\left(\dfrac{4}{x^2}\right)}$

$\sf{\implies\,\dfrac{d^2y}{dx^2}=y\left\{2+\dfrac{8}{x^2}+\dfrac{8}{x}-\dfrac{4}{x^2}\right\}}$

$\sf{\implies\,\dfrac{d^2y}{dx^2}=y\left(2+\dfrac{4}{x^2}+\dfrac{8}{x}\right)}$

$\sf{\implies\,\dfrac{d^2y}{dx^2}=2y\left(1+\dfrac{2}{x^2}+\dfrac{4}{x}\right)}$

$\sf{\implies\,\dfrac{d^3y}{dx^3}=2\left(1+\dfrac{2}{x^2}+\dfrac{4}{x}\right)\dfrac{dy}{dx}+2y\left(-\dfrac{4}{x^3}-\dfrac{4}{x^2}\right)}$

$\sf{\implies\,\dfrac{d^3y}{dx^3}=4y\left(1+\dfrac{2}{x}\right)\left(1+\dfrac{2}{x^2}+\dfrac{4}{x}\right)+4y\left(-\dfrac{2}{x^3}-\dfrac{2}{x^2}\right)}$

$\sf{\implies\,\dfrac{d^3y}{dx^3}=4y\left\{\left(1+\dfrac{2}{x}\right)\left(1+\dfrac{2}{x^2}+\dfrac{4}{x}\right)+\left(-\dfrac{2}{x^3}-\dfrac{2}{x^2}\right)\right\}}$

$\sf{\implies\,\dfrac{d^3y}{dx^3}=4y\left\{1+\dfrac{2}{x^2}+\dfrac{4}{x}+\dfrac{2}{x}+\dfrac{4}{x^3}+\dfrac{8}{x^2}-\dfrac{2}{x^3}-\dfrac{2}{x^2}\right\}}$

$\sf{\implies\,\dfrac{d^3y}{dx^3}=4y\left\{1+\dfrac{4}{x}+\dfrac{2}{x}+\dfrac{4}{x^3}+\dfrac{8}{x^2}-\dfrac{2}{x^3}\right\}}$

$\sf{\implies\,\dfrac{d^3y}{dx^3}=4y\left\{1+\dfrac{6}{x}+\dfrac{2}{x^3}+\dfrac{8}{x^2}\right\}}$

$\sf{\implies\,\dfrac{d^3y}{dx^3}=4\cdot\,x^4\cdot\,e^{2x}\left\{1+\dfrac{6}{x}+\dfrac{2}{x^3}+\dfrac{8}{x^2}\right\}}$

$\sf{\implies\,\dfrac{d^3y}{dx^3}=4\cdot\,e^{2x}\left\{x^4+6x^3+2x+8x^2\right\}}$

Answer 2

Step-by-step explanation:

Answer:

Step-by-step explanation:

We have,

\sf{y=x^4\,e^{2x}}y=x

4

e

2x

\sf{\implies\,\dfrac{dy}{dx}=4x^3\,e^{2x}+2x^4\,e^{2x}}⟹

dx

dy

=4x

3

e

2x

+2x

4

e

2x

\sf{\implies\,\dfrac{dy}{dx}=2x^4\,e^{2x}\left(\dfrac{2}{x}+1\right)}⟹

dx

dy

=2x

4

e

2x

(

x

2

+1)

\sf{\implies\,\dfrac{dy}{dx}=2y\left(1+\dfrac{2}{x}\right)}⟹

dx

dy

=2y(1+

x

2

)

\sf{\implies\,\dfrac{d^2y}{dx^2}=2\left(1+\dfrac{2}{x}\right)\dfrac{dy}{dx}+2y\left(-\dfrac{2}{x^2}\right)}⟹

dx

2

d

2

y

=2(1+

x

2

)

dx

dy

+2y(−

x

2

2

)

\sf{\implies\,\dfrac{d^2y}{dx^2}=2y\left(1+\dfrac{2}{x}\right)^2-y\left(\dfrac{4}{x^2}\right)}⟹

dx

2

d

2

y

=2y(1+

x

2

)

2

−y(

x

2

4

)

\sf{\implies\,\dfrac{d^2y}{dx^2}=y\left\{2+\dfrac{8}{x^2}+\dfrac{8}{x}-\dfrac{4}{x^2}\right\}}⟹

dx

2

d

2

y

=y{2+

x

2

8

+

x

8

−

x

2

4

}

\sf{\implies\,\dfrac{d^2y}{dx^2}=y\left(2+\dfrac{4}{x^2}+\dfrac{8}{x}\right)}⟹

dx

2

d

2

y

=y(2+

x

2

4

+

x

8

)

\sf{\implies\,\dfrac{d^2y}{dx^2}=2y\left(1+\dfrac{2}{x^2}+\dfrac{4}{x}\right)}