I'm missing a step somewhere on this: $$f(t)=e^{t\sin2t}.$$
I know I have to use the chain rule but I'm getting tripped up. here are my steps:
$\frac d{dx}(e^{t\sin2t})$
$e^{t\sin2t} \frac d{dx}(t\sin2t)$ which then becomes a product rule:
$e^{t\sin2t} [\frac d{dx}(t) (\sin2t) + \frac d{dx}(\sin2t) (t)]$
then the $\sin 2t$ needs the chain rule:
$(\sin2t) \frac d{dx}(2t)$ the results of this are $2t\sin2t$ which I plug back in to the above equation:
$e^{t\sin2t} [\frac d{dx}(t) (\sin2t) + (t) (2t\sin2t)]$ which gives me:
$e^{t\sin2t} [(t\sin2t) + (2t^{2}\sin2t)]$
I get lost from here. The answer says its:
$F(t)=e^{t\sin2t}(2t\cos2t +\sin2t)$
$\endgroup$ 42 Answers
$\begingroup$You didn't use product rule correctly
$(t\cdot \sin{2t})'=t'\sin{2}t+t(\sin{2}t)'=1\cdot\sin{2t}+t\cdot \cos{2t}\cdot 2$
because
$(\sin{2t})'=\cos{2t} \cdot(2t)'=\cos2t \cdot2$
So you get:
$\endgroup$ 2 $\begingroup$$e^{t\sin2t}(\sin{2t}+2t\cos{2t})$
By the Chain Rule,
$$\frac{d}{dt}\left(e^{t\sin{2t}}\right) = e^{t\sin{2t}}\cdot\frac{d}{dt}\left(t\sin{2t}\right)$$
But by the product rule,
$$\begin{align}\frac{d}{dt}\left(t\sin{2t}\right) &= t\cdot\frac{d}{dt}(\sin2t) + \sin2t\cdot\frac{d}{dt}(t)\\ &=t\cdot2\cos2t + \sin2t\\ &= 2t\cos2t + \sin2t\end{align}$$
Hence
$$\frac{d}{dt}\left(e^{t\sin{2t}}\right) = e^{t\sin{2t}}(2t\cos2t + \sin2t)$$
$\endgroup$
