Finding the tenth derivative of $f(x) = e^x\sin x$ at $x=0$ I came across this Question where I have to find $$f^{(10)}$$ for the following function at $x = 0$
$$f(x) = e^x\sin x$$
I tried differentiating a few times to get a pattern but didn’t get one, can someone provide the solution.
 A: Hint:
As $\;\mathrm e^x\sin x=\operatorname{Im}\bigl(\mathrm e^{(1+i)x}\bigr)$, you have to find first the real and imaginary parts of $(1+i)^{10}$.
Some details:
There results from the above remark  and linearity of differentiation that $\;(\mathrm e^x\sin x)'=\bigl(\operatorname{Im}(\mathrm e^{(1+i)x})\bigr)'= \operatorname{Im}\bigl((1+i)\mathrm e^{(1+i)x}\bigr)$, hence
$$\;(\mathrm e^x\sin x)''=\bigl(\operatorname{Im}((1+i)\mathrm e^{(1+i)x}))\bigr)'= \operatorname{Im}\bigl((1+i)^2\mathrm e^{(1+i)x}\bigr),$$
and more generally
$$(\mathrm e^x\sin x)^{(k)}=\bigl(\operatorname{Im}(\mathrm e^{(1+i)x})\bigr)^{(k)}=\operatorname{Im}\bigl((1+i)^k(\mathrm e^{(1+i)x})\bigr).$$
A: Hint:
$$f(x)=e^x\sin x$$
$$f'(x)=e^x(\sin x +\cos x)$$
$$f''(x)=e^x(\sin x+\cos x)+e^x(\cos x -\sin x)=2e^x(\cos x)$$
$$f'''(x)=e^x(2\cos x)-e^x(2\sin x)=2e^x(\cos x-\sin x)$$
$$f^{IV}(x)=2e^x(\cos x-\sin x)-2e^x(\cos x+\sin x)=-4e^x(\sin x)=-4f(x)$$
A: One trick here is to use $e^{ix}=\cos x+i\sin x$ and define $g(x)=e^x\cos x$ then $f(x)$ and $g(x)$ are both real functions. 
Let $h(x)=g(x)+if(x)=e^{(1+i)x}$ then the tenth derivative of $h(x)$ is $(1+i)^{10}h(x)$ and the tenth derivative of $f(x)$ is the imaginary part of this.
Because you only want the value at $x=0$ you can evaluate there, with $h(0)=g(0)+if(0)$
A: $$(e^x(a\cos x+b\sin x))'=e^x(a\cos x+b\sin x)+e^x(b\cos x-a\sin x)=e^x((a+b)\cos x+(b-a)\sin x).$$
So
$$(0,1)\to(1,1)\to(2,0)\to(2,-2)\to(0,-4)\to(-4,-4)\to(-8,0)\to(-8,8)\to(0,16)\to(16,16)\to(32,0).$$

If you divide by increasing powers of $2$, in pairs, the pattern emerges, with period $8$:
$$(0,1)\to(1,1)\to(1,0)\to(1,-1)\to(0,-1)\to(-1,-1)\to(-1,0)\to(-1,1)\to(0,1)\to(1,1)\to(1,0).$$
A: I get $f^{10}(x)=32e^x\cos x$.
Here's what I did:
\begin{align}f'(x)&=e^x(\sin x+\cos x)\\
\implies f''(x)&=2e^x\cos x\\
\implies f^3(x)&=2e^x(\cos x-\sin x)\\
\implies f^4(x)&=2e^x(-2\sin x)=-4f(x)\\
\implies f^5(x)&=-4f'(x)\\
\implies f^8(x)&=-4f^4(x)=16f(x)\\
\implies f^{10}(x)&=16f''(x)\end{align}
A: Regarding the question about looking for a pattern:
Repeated application of the product rule gives
$$ f(x) = e^x \sin x$$
$$ f'(x) = e^x \sin x + e^x \cos x$$
$$ f''(x) = e^x \sin x + e^x \cos x + e^x \cos x - e^x \sin x = 2\ e^x \cos x$$
Is a pattern emerging?
$$ f^{(3)}(x) = 2\ e^x \cos x -2\ e^x \sin x$$
$$ f^{(4)}(x) = 2\ e^x \cos x -2\ e^x \sin x - 2\ e^x \sin x - 2\ e^x \cos x = -4\ e^x \sin x$$
Yes. We can then conclude that
$ f^{(6)}(x) = -8\ e^x \cos x$, $f^{(8)}(x) = 16\ e^x \sin x$, and $f^{(10)}(x) = 32\ e^x \cos x$ 
such that $f^{(10)}(0) = 32 $
A: Use the formula $(fg)^{(n)}= \sum\limits_{k=0}^{n} \binom {n} {k} (f)^{(k)}(g)^{(n-k)}$. 
A: Using recurrence relation:
$$\begin{align}f^{(0)}=&e^x\sin x\\
f^{(1)}=&e^x\sin x+e^x\cos x=f^{(0)}+e^x\cos x\\
f^{(2)}=&f^{(1)}+\color{red}{e^x\cos x}-e^x\sin x=f^{(1)}+\color{red}{f^{(1)}-f^{(0)}}-f^{(0)}\\
\color{blue}{f^{(n)}=}&\color{blue}{2f^{(n-1)}-2f^{(n-2)}, f^{(0)}(0)=0, f^{(1)}(0)=1} \Rightarrow \\
f^{(n)}=&\frac12i\left[(1-i)^n-(1+i)^n\right] \Rightarrow \\
f^{(10)}(0)=&\frac12i[(1-i)^{10}-(1+i)^{10}]=\\
=&-\frac12i\left[{10\choose 1}i+{10\choose 3}i^3+{10\choose 5}i^5+{10\choose 7}i^7+{10\choose 9}i^9\right]=\\
=&{10\choose 1}-{10\choose 3}+{10\choose 5}-{10\choose 7}+{10\choose 9}=\\
=&10-120+252-120+10=\\
=&32.\end{align}$$

Addendum: Direct calculation from the recurrence relation above:
$$\color{blue}{f^{(n)}=2\left[f^{(n-1)}-f^{(n-2)}\right], f^{(0)}(0)=0, f^{(1)}(0)=1}\\
\begin{array}{c|c|c|c|c|c|c|c|c}
f^{(0)}&f^{(1)}&f^{(2)}&f^{(3)}&f^{(4)}&f^{(5)}&f^{(6)}&f^{(7)}&f^{(8)}&f^{(9)}&f^{(10)}\\
\hline
0&1&2&2&0&-4&-8&-8&0&16&32\end{array}$$
A: Using power series: it is well-known that $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$ and $\sin(x)=\sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{(2k+1)!}$ for any real number $x$, so 
$$ e^x\sin(x)=(1+x+\frac{x^2}{2!}+\dots+\frac{x^{10}}{10!}+\dots)(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\dots) $$
By expanding, the coefficient of $x^{10}$ is $\frac{1}{9!1!}-\frac{1}{7!3!}+\frac{1}{5!5!}-\frac{1}{7!3!}+\frac{1}{9!1!}$
But this coefficient is also $\frac{f^{(10)}(0)}{10!}$, so 
$$ f^{(10)}(0)=\frac{10!}{9!1!}-\frac{10!}{7!3!}+\frac{10!}{5!5!}-\frac{10!}{7!3!}+\frac{10!}{9!1!} = 10 -120 + 252 - 120 +10 = 32$$
A: Use Leibniz' Rule for higher derivatives of a product:
$$\frac{d^n}{dx^n}(uv)
  =\frac{d^nu}{dx^n}v+\binom n1\frac{d^{n-1}u}{dx^{n-1}}\frac{dv}{dx}
    +\binom n2\frac{d^{n-2}u}{dx^{n-2}}\frac{d^2v}{dx^2}+\cdots+u\frac{d^nv}{dx^n}\ .$$
In your case take $u=e^x$ and $v=\sin x$.  Since you are going to substitute $x=0$ after differentiating, all the $e^x$ terms will be $1$, all the $\sin x$ terms will be $0$ and all the cos $x$ terms will be $1$ (though some of them will pick up a negative sign when you differentiate).  So the answer is
$$\eqalign{0+\binom{10}11&{}+\binom{10}20+\binom{10}3(-1)+\binom{10}40+\binom{10}51\cr
  &\qquad{}+\binom{10}60+\binom{10}7(-1)+\binom{10}80+\binom{10}91+\binom{10}{10}0\cr
  &=10-120+252-120+10\cr
  &=32\ .\cr}$$
