Question

please help me do this question​

Answer 1

Answer:

$\dfrac{d^2y}{dx^2} = 5e^{3x} \sin (2x)+12e^{3x} \cos (2x)$

Step-by-step explanation:

$\large \boxed{\begin{minipage}{5 cm}\textsf{\underline{Product Rule}}\\\\ \textsf{When } \:y=uv\\\\\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v \dfrac{du}{dx} \\\end{minipage}}$

Given:

$y=e^{3x} \sin(2x)$

First Derivative

$\textsf{let } u=e^{3x} \implies \dfrac{du}{dx}=3e^{3x}$

$\textsf{let } v= \sin(2x) \implies \dfrac{dv}{dx}=2 \cos (2x)$

Using the Product Rule:

$\begin{aligned}\dfrac{dy}{dx} & =e^{3x}\cdot 2 \cos (2x) + \sin (2x) \cdot 3e^{3x}\\& = 2e^{3x} \cos (2x) + 3e^{3x} \sin (2x)\\& = e^{3x}(2 \cos (2x) + 3 \sin (2x))\end{aligned}$

Second Derivative

$\textsf{let } u=e^{3x} \implies \dfrac{du}{dx}=3e^{3x}$

$\textsf{let } v= 2 \cos (2x)+3 \sin (2x) \implies \dfrac{dv}{dx}=-4 \sin (2x) +6 \cos (2x)$

Using the Product Rule:

$\begin{aligned}\dfrac{d^2y}{dx^2} & =e^{3x}(-4 \sin (2x) +6 \cos (2x)) +3e^{3x} (2 \cos (2x)+3 \sin (2x))\\& = e^{3x}[-4 \sin (2x) +6 \cos (2x)+6 \cos (2x)+9 \sin (2x)]\\& = e^{3x}[5 \sin (2x) +12 \cos (2x)]\\& = 5e^{3x} \sin (2x)+12e^{3x} \cos (2x)\end{aligned}$

Answer 2

Answer:

$\tt{ \frac{ {d}^{2} y}{d {x}^{2} } = 5 {e}^{3x} \sin(2x) + 12 {e}^{3x} \cos(2x)}$

Step-by-step explanation:

⠀

⠀

⠀⠀⠀⠀▩ Question

$y = {e}^{3x} \sin(2x) \\ { \sf{To \: find}} \: \frac{ {d}^{2}y }{d {x}^{2} }$

⠀

⠀

⠀⠀⠀⠀■ Solution

⠀

• First finding dy/dx

⠀

$\sf using \: \frac{d}{dx} uv = u \frac{d}{dx} v + v \frac{d}{dx} u \: formula$

⠀

⠀

• Taking derivative of sin2x and e^3x

⠀

$\bold{ \frac{dy}{dx} = {e}^{3x} \frac{d}{dx} \sin(2x) + \sin(2x) \frac{d}{dx} {e}^{3x} } \\ \\ \bold{ \frac{dy}{dx} = {e}^{3x} \cos(2x) \frac{d}{dx} 2x + \sin(2x) {e}^{3x} \frac{d}{dx} 3x}$

⠀

⠀

• Multiplying

⠀

$\bold{ \frac{dy}{dx} = {e}^{3x} \times 2 \cos(2x) + \sin(3x) \times 3 {e}^{3x} }$

⠀

⠀

• Value of dy/dx

⠀

$\bold{ \frac{dy}{dx} = 2 {e}^{3x} \cos(2x) + 3 {e}^{3x} \sin(2x) }$

⠀

⠀

• Now taking derivative again using same formula as mention in first part, so that to find d²y/dx².

⠀

⠀

⠀⠀● First solving 2e^3x cos(2x)

⠀

$\bold{ 2( {e}^{3x} \frac{d}{dx} \cos(2x) + \cos(2x) \frac{d}{dx} {e}^{3x}) }$

⠀

⠀

• Taking derivative of cos2x and e^3x

⠀

$\bold{2( {e}^{3x} - \sin(2x) \frac{d}{dx} 2x + \cos(2x) {e}^{3x} \frac{d}{dx} 3x)} \\ \\ \bold{ - 4 {e}^{3x} \sin(2x) + 6 {e}^{3x} \cos(2x) }$

⠀

⠀

⠀⠀● Now solving 3e^3x sin(2x)

⠀

$\bold{ 3( {e}^{3x} \frac{d}{dx} \sin(2x) + \sin(2x) \frac{d}{dx} {e}^{3x} )}$

⠀

⠀

• Now taking derivative of sin 2x and e^3x

⠀

$\bold{ 3( {e}^{3x} \cos(2x) \frac{d}{dx} 2x + \sin(2x) {e}^{3x} \frac{d}{dx} 3x)} \\ \\ \bold{ 6 {e}^{3x} \cos(2x) + 9 {e}^{3x} \sin(2x) }$

⠀

⠀

• Adding both solution

⠀

$\bold{ \frac{ {d}^{2}y }{d {x}^{2} } = - 4 {e}^{3x} \sin(2x) + 6 {e}^{3x} \cos(2x) + 6 {e}^{3x} \cos(2x) + 9 {e}^{3x} \sin(2x) } \\ \\ \bold{ \frac{ {d}^{2} y}{d {x}^{2} } = 5 {e}^{3x} \sin(2x) + 12 {e}^{3x} \cos(2x) }$