How do you differentiate this...?

Question

$e^{t sin 2 t}$ $e^(tsin2t)$

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Answer 1 · 2018-06-10T07:08:04

$\frac{d y}{d t} = (sin (2 t) + 2 t cos (2 t)) e^{t \cdot sin (2 t)}$ $dy/dt=(sin(2t)+2tcos(2t))e^(t*sin(2t)$

We want to find the derivative of

$y = e^{t \cdot sin (2 t)}$ $y=e^(t*sin(2t))$

By the chain rule we have

$color(blue)(y=e^(f(x))=>dy/dx=f'(x)e^(f(x))$

So for your problem $color(red)(f(t)=t*sin(2t)$ , and by the product rule

$color(red)(f'(t)=sin(2t)+2tcos(2t)$

Thus

$dy/dt=(sin(2t)+2tcos(2t))e^(t*sin(2t)$