# Math Help - how to differentiate

1. ## how to differentiate

Using an appropriate rule how would I differentiate

y = (4x2 - e2x) sin 3x

2. ## Re: how to differentiate

Originally Posted by tomjay
Using an appropriate rule how would I differentiate

y = (4x2 - e2x) sin 3x

The product rule is the only appropriate rule.

3. ## Re: how to differentiate

The Rules needed are Product, Difference, Chain and Power.

4. ## Re: how to differentiate

Hello, tomjay!

Since you're working with exponential and trigonometric functions,
. . you MUST be familiar with the Product Rule . . . right?

$\text{Differentiate: }\:y \:=\: (4x^2 - e^{2x})\sin 3x$

$\text{Product Rule: \;If }y \:=\:f(x)\!\cdot\!g(x),\,\text{ then: }\:y' \;=\;f'(x)\!\cdot\!g(x) + f(x)\!\cdot\!g'(x)$

$\text{We have: }\:y \;=\;\overbrace{(4x^2-e^{2x})}^{f(x)}\cdot\overbrace{\sin3x}^{g(x)}$

$\text{Then: }\:y' \;=\;\overbrace{(8x- 2e^{2x})}^{f'(x)}\cdot\overbrace{\sin3x}^{g(x)} \;+\; \overbrace{(4x^2-e^{2x}) }^{f(x)}\cdot\overbrace{3\cos3x}^{g'(x)}$

5. ## Re: how to differentiate

Originally Posted by Soroban
Hello, tomjay!

Since you're working with exponential and trigonometric functions,
. . you MUST be familiar with the Product Rule . . . right?

$\text{Product Rule: \;If }y \:=\:f(x)\!\cdot\!g(x),\,\text{ then: }\:y' \;=\;f'(x)\!\cdot\!g(x) + f(x)\!\cdot\!g'(x)$

$\text{We have: }\:y \;=\;\overbrace{(4x^2-e^{2x})}^{f(x)}\cdot\overbrace{\sin3x}^{g(x)}$

$\text{Then: }\:y' \;=\;\overbrace{(8x- 2e^{2x})}^{f'(x)}\cdot\overbrace{\sin3x}^{g(x)} \;+\; \overbrace{(4x^2-e^{2x}) }^{f(x)}\cdot\overbrace{3\cos3x}^{g'(x)}$
That's assuming that \displaystyle \begin{align*} e^{2x} \end{align*} is the correct function. As written, it should be \displaystyle \begin{align*} e^2x \end{align*}...