Find the nth Derivative of $e^{2x}\sin x\sin2x$

Question

I need to find the $n$-th derivative of $e^{2x}\sin x\sin2x$. So anyway first we will devide the qustion into two parts u and v $$y=uv$$ $$u=e^2x$$ $$v=sinxsin2x$$ now $$sinasinb=\dfrac{1}2[cos(a−b)−cos(a+b)]$$ then subsitute: $$sinasinb=\dfrac{1}2e^2x[cos(1−2)x−cos(1+2)x]$$ btw $$cos(-x)=cos(x)$$ so now we get $$sinasinb=\dfrac{1}2e^2x[cos(x)−cos(3x)]$$ so we bring the $$e^2x$$ inside and get $$sinasinb=\dfrac{1}2[e^2x.cos(x)−e^2x.cos(3x)]$$ now we use another formula $$y=e^{ax}.cosx(bx+c)$$ $$yn=(a^2+b^2)^{n/2}.e^{ax}.cos(bx+c+ntan^{-1}\dfrac{b}a$$ selfnote:find the command for subscript and edit it in)

for part A:

a=2,b=1,c=0 which gives $$\dfrac{1}2(5^\dfrac{n}2.e^{2x}.cos(x+ntan^{-1}\dfrac{1}2))$$

for part B

a=2,b=3,c=0 which gives $$\dfrac{1}2(13^\dfrac{n}2.e^{2x}.cos(3x+ntan^{-1}\dfrac{3}2))$$

now we combine them both to get

$$\dfrac{1}2(5^\dfrac{n}2.e^{2x}.cos(x+ntan^{-1}\dfrac{1}2))-(13^\dfrac{n}2.e^{2x}.cos(3x+ntan^{-1}\dfrac{3}2))$$ if theres a way to improve my answer pls tell me

Do you need to find the $n$th derivative or the $n$th derivative at a point? — Ninad Munshi, Dec 29 '22 at 11:53
If the question is asking for general answers, then start off by finding a pattern in the derivatives. Ie. how many "derivatives" later does $\sin{x}$ and $e^x$ repeat? — Dstarred, Dec 29 '22 at 11:57
You can also transform the product of sine functions in this way: $\sin a \sin b=\frac12[\cos(a-b)-\cos(a+b)]$ — Jean Marie, Dec 29 '22 at 12:01
There is also the trinomial theorem and very similar formula for general Leibniz rule for more than 2 factors — Тyma Gaidash, Dec 29 '22 at 12:23

score 3 · Accepted Answer · edited Dec 29 '22 at 12:20

Hint:

Use the transformation $\sin\alpha\cdot\sin\beta=-\frac{1}{2}\left[\cos(\alpha+\beta)-\cos(\alpha-\beta)\right]$.

Meanwhile, notice that $g(x)=f(x)e^{2x},g'(x)=e^{2x}(f'(x)+2f(x))$, the $n$-th derivative of the given function should be like $h^{(n)}(x)=-\frac{1}{2}e^{2x}(A\cos 3x+B\cos x+C\sin x+D\sin 3x)$. All you need to find is $A, B, C, D$.

Another acceptable method is, as @Empy2 mentioned above, using the equation below: $$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$

Find the nth Derivative of $e^{2x}\sin x\sin2x$

1 Answers1