Find the $n$ derivative of $y= e^{2x}\sin^2 x$ We have 
\begin{align*}
y&= e^{2x}\sin^2 x\\
&= e^{2x}\left(\frac{1-\cos 2x}{2}\right)\\
&= \frac{e^{2x}}{2} - \frac{e^{2x}\cos 2x}{2}
\end{align*}
Then
\begin{align*}
y^{(n)} &= \left(\frac{e^{2x}}{2}\right)^{(n)} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}\\
&= 2^{n-1}e^{2x} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}
\end{align*}
I don't know how to proceed with the rightmost term. So far I've been applying the Leibniz Rule whenever I've had to find the $n$ derivative of a function of the form $f(x)g(x)$, because is clear that either $f(x)$ or $g(x)$ has a derivative a $k$ derivative ($1<k<n$) equal to zero, which simplifies the expression nicely.
But here
$$\frac{e^{2x}\cos 2x}{2}$$
both functions are infinitely differentiable on $\mathbb{R}$ which makes things a bit different.
My only attempt was to write its n derivative in this form
\begin{align*}
&=\frac{1}{2}\sum_{k=0}^n{n \choose k} \big(2^k e^{2x}\big)\bigg(2^{n-k}\cos \left[2x + \frac{\pi(n-k)}{2}\right]\bigg)\\
&=\sum_{k=0}^n {n \choose k} 2^{n-1}e^{2x}\cos \left[2x + \frac{\pi(n-k)}{2}\right]
\end{align*}
So 
\begin{align*}
y^{(n)} &= 2^{n-1}e^{2x} -\sum_{k=0}^n{n \choose k} 2^{n-1}e^{2x}\cos \left[2x + \frac{\pi(n-k)}{2}\right]\\
&= 2^{n-1}e^{2x}\left(1 -\sum_{k=0}^n{n \choose k} \cos \left[2x + \frac{\pi(n-k)}{2}\right]\right)
\end{align*}
But the textbook's answer is
$$2^{n-1}e^{2x}\left(1 -2^{n/2}\cos \left[2x + \frac{\pi n}{4}\right]\right)$$
For some reason I have the feeling that a little of modular arithmetic has to be applied on $\frac{\pi(n-k)}{2}$
 A: You're correct that Leibniz's Rule is a sound way forward.  Note that we have
$$\begin{align}
e^{2x}\sin^2(x)=\frac12 e^{2x}-\frac12 e^{2x}\cos(2x)
\end{align}$$
Then, taking the $n$'th order derivative, we have
$$\begin{align}
\frac{d^n}{dx^n}\left(e^{2x}\sin^2(x)\right)&=\frac{d^n}{dx^n}\left(\frac12 e^{2x}-\frac12 e^{2x}\cos(2x)\right)\\\\
&=2^{n-1}e^{2x}-\frac12 \sum_{k=0}^n\binom{n}{k}\frac{d^{n-k}e^{2x}}{dx^{n-k}}\frac{d^k\cos(2x)}{dx^k}\\\\
&=2^{n-1}e^{2x}\left(1-\sum_{k=0}^n\binom{n}{k}\frac{d^k\cos(2x)}{d(2x)^k}\right)\\\\
\end{align}$$
So, the problem boils down to taking the $k$'th derivative of the cosine function.  But we can express that derivative as $\cos(x+k\pi/2)$.  Hence, we have
$$\begin{align}
\frac{d^n}{dx^n}\left(e^{2x}\sin^2(x)\right)&=2^{n-1}e^{2x}\left(1-\sum_{k=0}^n\binom{n}{k}\cos(x+k\pi/2)\right)\\\\
&=2^{n-1}e^{2x}\left(1-\text{Re}\left(\sum_{k=0}^n\binom{n}{k}e^{i(x+k\pi/2)}\right)\right)\\\\
&=2^{n-1}e^{2x}\left(1-\text{Re}\left(e^{ix}(1+i)^n\right)\right)\\\\
&=2^{n-1}e^{2x}\left(1-\text{Re}\left(2^{n/2}e^{i(x+n\pi/4)}\right)\right)\\\\
&=2^{n-1}e^{2x}\left(1-2^{n/2}\cos(x+n\pi/4)\right)\\\\
\end{align}$$
as was to be shown!
A: We know that $$\sin^2(x) =  \dfrac{1 - \cos(2x)}{2}= \dfrac{1}{2} -\dfrac{1}{2}\Re(\exp(2ix)), $$ where $\Re$ is the real part of a complex number.
It follows that
$$y =\exp(2x)\sin^2(x) = \dfrac{\exp(2x)}{2} - \dfrac{1}{2}\Re(\exp(zx)), $$ with $z = 2+2i.$  
Using complex-valued derivatives:
$$\dfrac{dy}{dx} = \exp(2x)- \dfrac{z}{4}\exp(zx)- \dfrac{z^{*}}{4}\exp(z^{*}x),$$ which leads to
$$\dfrac{d^ny}{dx^n} = 2^{n-1}\exp(2x) - \dfrac{z^n}{4}\exp(zx) - \dfrac{z^{*n}}{4}\exp(z^{*}x) = 2^{n-1}\exp(2x) - \dfrac{1}{2}\Re (z^n \exp(zx)).$$
A: Calculate the fist few derivatives and see if a recognizable pattern shows up.
$f(x) = \frac {e^{2x}}{2} + \frac {e^{2x}\cos2x}{2}\\
f'(x) = e^{2x} + e^{2x} \cos 2x - e^{2x}\sin 2x\\
f''(x) = 2e^{2x} - 2e^{2x}\cos 2x - 2e^{2x}\sin 2x - 2e^{2x}\sin 2x - 2e^{2x} \cos 2x = 2e^{2x} - 4e^{2x}\sin 2x\\
f'''(x) = 4e^{2x} - 8 e^{2x}\sin 2x - 8 e^{2x}\cos 2x\\
f^{(4)} = 8e^{2x} - 16\cos 2x$
Do you see a pattern? Can you prove that the pattern will continue?
If you know a little bit about complex exponential you could say:
$f(x) = \frac {e^{2x}}{2} + Re[\frac {e^{(2+2i)x}}{2}]\\
f^{(n)}(x) = 2^n\frac {e^{2x}}{2} + \frac {(2\sqrt2)^n}{2} Re[\frac {e^{(2+2i)x + \frac {n\pi}{4}i}}{4}]\\
f^{(n)}(x) = 2^n\frac {e^{2x}}{2} + \frac {(2\sqrt2)^n}{2} e^{2x} \cos(2x + \frac {n\pi}{4})\\$
A: Taking into consideration just the second part, let's call 
$$p(x)=e^{2x}\cos 2x$$
then 
\begin{align}
&p^{(1)}(x)=2e^{2x}\cos 2x-2e^{2x}\sin 2x\\
&p^{(1)}(x)=2e^{2x}(\cos 2x-\sin 2x)=2e^{2x}\sqrt{2}\left(\frac{\sqrt{2}}{2}\cos 2x-\frac{\sqrt{2}}{2}\sin 2x\right)\\
&p^{(1)}(x)=2^{3/2}e^{2x}\left(\cos(\pi/4)\cos 2x-\sin(\pi/4)\sin 2x\right)\\
&p^{(1)}(x)=2^{3/2}e^{2x}\cos\left( 2x+\pi/4 \right)
\end{align}
and following the same calculation that we just did we can conclude that 
$$p^{(2)}(x)=2^{3/2}2^{3/2}e^{2x}\cos\left( 2x+\pi/4+\pi/4 \right)$$
Now we can see the pattern. So we can prove by finite induction that 
$$p^{(n)}(x)=2^{3n/2}e^{2x}\cos\left( 2x+n\pi/4\right)$$
and finaly:
$$\left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}=2^{3n/2-1}e^{2x}\cos\left( 2x+n\pi/4\right)$$
A: The crux of the problem boils down to computing the $n$th derivative of the function $f(u)=e^u\cos u$.  (Note that if $g(x)=f(2x)$, then $g^{(n)}(x)=2^nf^{(n)}(2x)$.)  Using $\cos u=\Re(e^{iu})$ and $1+i=\sqrt2e^{i\pi/4}$, we find
$$f(u)=\Re(e^{(1+i)u})\implies f^{(n)}(u)=\Re((1+i)^ne^{(1+i)u})=2^{n/2}e^u\Re(e^{(u+n\pi/4)i})=2^{n/2}e^u\cos\left(u+{n\pi\over4}\right)$$
Applied to the OP's problem, we have
$$y(x)={e^{2x}\over2}-{f(2x)\over2}\implies y^{(n)}(x)={2^ne^{2x}\over2}-{2^nf^{(n)}(2x)\over 2}=2^{n-1}e^{2x}\left(1-2^{n/2}\cos\left(2x+{n\pi\over4} \right)\right)$$
