How do you find the first and second derivative of y=e^(alphax)sinbetax?

1 Answer
Feb 7, 2017

$y = {e}^{\alpha x} \sin \left(\beta x\right)$

$\implies$

$y ' ' = {\alpha}^{2} {e}^{\alpha x} \sin \left(\beta x\right) + 2 \alpha \beta {e}^{\alpha x} \cos \left(\beta x\right) - {\beta}^{2} {e}^{\alpha x} \sin \left(\beta x\right)$

Explanation:

First, remember the product rule states

$\left(f \left(x\right) g \left(x\right)\right) ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

In this case

$f \left(x\right) = {e}^{\alpha x}$

and

$g \left(x\right) = \sin \left(\beta x\right)$

In order to find the derivative of $y \left(x\right)$, using the product rule, we will first need to find $f ' \left(x\right)$ and $g ' \left(x\right)$.

To find these, we will need the chain rule.

(a(b(x))'=a'(b(x))b'(x)

In the case of $f \left(x\right)$

$a \left(x\right) = {e}^{x}$

and

$b \left(x\right) = \alpha x$

Since ${e}^{x}$ is its own derivative

$f ' \left(x\right) = \left(b ' \left(x\right)\right) {e}^{b \left(x\right)} = \left(\alpha x\right) ' {e}^{\alpha x}$

And since $\left(c x\right) ' = c \left(x\right) ' = c$

$f ' \left(x\right) = \alpha {e}^{\alpha x}$

and in the case of $g \left(x\right)$

$a \left(x\right) = \sin \left(x\right)$

and

$b \left(x\right) = \beta x$

Then, using the same rules from above we have

$g ' \left(x\right) = \left(\sin \left(\beta x\right)\right) ' = \left(\beta x\right) ' \sin ' \left(\beta x\right) = \beta \sin ' \left(\beta x\right)$

and since $\left(\sin \left(x\right)\right) ' = \cos \left(x\right)$

$g ' \left(x\right) = \beta \cos \left(\beta x\right)$

Then plugging in we get

$y = {e}^{\alpha x} \sin \left(\beta x\right) \implies \underline{y ' = \alpha {e}^{\alpha x} \sin \left(\beta x\right) + \beta {e}^{\alpha x} \cos \left(\beta x\right)}$

Now, we have the first derivative. So, to find the second derivative, we now take the derivative of $y ' \left(x\right)$.

Since $\left(f \left(x\right) + g \left(x\right)\right) ' = f ' \left(x\right) + g ' \left(x\right)$

Then

$y ' ' = \left(\alpha {e}^{\alpha x} \sin \left(\beta x\right)\right) ' + \left(\beta {e}^{\alpha x} \cos \left(\beta x\right)\right) '$

By above

$\left(\alpha {e}^{\alpha x} \sin \left(\beta x\right)\right) ' = \alpha \left({e}^{\alpha x} \sin \left(\beta x\right)\right) ' =$

$\alpha \left(\alpha {e}^{\alpha x} \sin \left(\beta x\right) + \beta {e}^{\alpha x} \cos \left(\beta x\right)\right)$

$= {\alpha}^{2} {e}^{\alpha x} \sin \left(\beta x\right) + \alpha \beta {e}^{\alpha x} \cos \left(\beta x\right)$

also by above

$\left(\beta {e}^{\alpha x} \cos \left(\beta x\right)\right) ' = \beta \left({e}^{\alpha x} \cos \left(\beta x\right)\right) '$

$= \beta \left(\alpha {e}^{\alpha x} \cos \left(\beta x\right) + {e}^{\alpha x} \cos ' \left(\beta x\right)\right)$

Since $\left(\cos \left(x\right)\right) ' = - \sin \left(x\right)$

$= \beta \left(\alpha {e}^{\alpha x} \cos \left(\beta x\right) - \beta {e}^{\alpha x} \sin \left(\beta x\right)\right)$

$= \alpha \beta {e}^{\alpha x} \cos \left(\beta x\right) - {\beta}^{2} {e}^{\alpha x} \sin \left(\beta x\right)$

Then we plug in

$y ' ' = {\alpha}^{2} {e}^{\alpha x} \sin \left(\beta x\right) + \alpha \beta {e}^{\alpha x} \cos \left(\beta x\right) + \alpha \beta {e}^{\alpha x} \cos \left(\beta x\right) - {\beta}^{2} {e}^{\alpha x} \sin \left(\beta x\right) = \underline{{\alpha}^{2} {e}^{\alpha x} \sin \left(\beta x\right) + 2 \alpha \beta {e}^{\alpha x} \cos \left(\beta x\right) - {\beta}^{2} {e}^{\alpha x} \sin \left(\beta x\right)}$