Question

Differentiate the function. F(t) = e^(2t sin 2t)

Answer 1

Hello!

Answer 2

Chain rule, do you know how to use it?

Answer 3

@xapproachesinfinity makes a good point!

I was hoping you were on before, but maybe it was a glitch.

Here's what you want to recognize:
You know how to differentiate \(e^{g(x)}\), and \(e^x\) is a function itself! So, you have a function within a function, and so you want to apply the chain rule.

For any two functions \(f(x)\) and \(g(x)\), the chain rule states that \[(f(g(x)))' = f'(g(x))\times g'(x)\]

Feel free to ask if you don't see how to apply it!

Answer 4

You have to use the chain rule for the power, but the power's chain rule will not be part of the power.

\(\Large\color{midnightblue}{ f(t) = e^{(2t~\sin 2t)} }\)
\(\Large\color{midnightblue}{ f~'(t) = e^{(2t~\sin 2t)} \times [2t(\frac{d}{dt}~\sin~2t)+\sin~2t(\frac{d}{dt}~2t)~]}\)
you are deriving the power by the product.

Now, the sin(2t) has another chain rule,
d/dx sin(2t) = cos(2t) * d/dt 2t = cos(2t) * 2=2cos(2t)

\(\Large\color{midnightblue}{ f~'(t) = e^{(2t~\sin 2t)} \times [2t(2\cos~2t)+\sin~(2t) ×(2)~]}\)

you can simplify this yourself....

Answer 5

Notice that in @SolomonZelman 's response, the exponent's derivative must be multiplied (because the chain rule says so). The product rule was applied immediately, so that's where that chunk to the right if the \(\large e^{(2t\ \sin2t)}\) comes from