Question

Find the derivative of the function:
\[f(t)=e ^{at} \sin bt\]

Answer 1

\[\large\rm f(t)=e^{at}\sin(bt)\]Start by setting up your product rule,\[\large\rm f'(t)=\color{royalblue}{\left[e^{at}\right]'}\sin(bt)+e^{at}\color{royalblue}{\left[\sin(bt)\right]'}\]

Answer 2

Okay, so it would be \[(a \times e ^{at})(sinbt)+(e^{at})(cosbt)\]

Answer 3

not quite

Answer 4

chain rule on sin derivative, missing a b

Answer 5

Oh okay (b cos bt)

Answer 6

factor out the exponent term, that looks like a familiar thing

Answer 7

Okay so it would be \[e ^{at}(a \sin bt + b \cos bt)\]

Answer 8

yeah, i think those are oscillation forms, damped, forced, from DE class, maybe not, been awhile

Answer 9

Ok, thanks! So if there's questions like this type, we would apply the product rule and then the chain rule? At first I was doing the chain rule and then things looked complicated.

Answer 10

depends , here you have 2 functions multiplied together

[f * g] ' = f * g' + f ' * g

do that first, the chain rule follows from that

Answer 11

Alright, thank you!

Answer 12

basically if you have a nested function you use chain rule,

here if you call the a*t some substitute variable like u, sin(u)

to get derivative with respect to t, you need
\[\frac{ d }{ dt }= \frac{ d }{ du }*\frac{ du }{ dt }\]

Answer 13

Ah okay, that makes sense. I'm fairly new at the chain rule which is why it seems complicated at the moment.

Answer 14

right, i think if you take time to write all the notation out until it is second nature, it is easier

Answer 15

gl

Answer 16

Thanks!