Question

Find y' and y''. y = eαx sin βx

Answer 1

Product rule :)

$\rm y=e^{\alpha x} sin{\beta x}$

Start by "setting up" your product rule,
makes it easier to manage,

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

Derivative of exponential gives you the same function back but with chain rule applied,

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

and derivative of sine gives us cosine, again with chain rule giving us an extra coefficient,

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

Factoring the exponential out gives us a nicer looking answer,

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

For our second derivative, same process but it will be a little bit more work this time. We'll start by "setting up" our product rule,

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

and differentiating where it's needed,

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

and again factoring out the exponential, distributing the alpha to the first two terms, and simplifying where possible,

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

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

any confusion along the way?