What will be the 49th derivative? Let be the function $f=(x^3+3x) \cdot sin(x)$
$f'(x)=\left(3x^2+3\right)\sin\left(x\right)+\left(x^3+3x\right)\cos\left(x\right)$
$f''(x)=\left(3x-x^3\right)\sin\left(x\right)+\left(6x^2+6\right)\cos\left(x\right)$
$f'''(x)=\left(-9x^2-3\right)\sin\left(x\right)+\left(15x-x^3\right)\cos\left(x\right)$
$f''''(x)=\left(x^3-33x\right)\sin\left(x\right)+\left(12-12x^2\right)\cos\left(x\right)$
I still can't the exact pattern in the results, I would appreciate any help.
 A: Going off A.Riesen's comment, the General Leibniz Rule states that
$$(uv)^{(n)} = \sum_{k=0}^n {n \choose k} u^{(k)} v^{(n-k)} %seriously, did nobody notice that I had written down the formula incorrectly the first time?$$
where $u,v$ are functions of $x$ and $f^{(n)}$ is the $n^{th}$ derivative of $f$ with respect to $x$. Plugging in $u = x^3 + 3x$, $v = \sin x$ and $n = 49$, we get
$$((x^3+3x)(\sin x))^{(49)} = \sum_{k=0}^{49} {n \choose k} (x^3+3x)^{(k)}(\sin x)^{(49-k)}$$
Fortunately, since $(x^3+3x)^{(n)} = 0$ for all $n \ge 4$, we can reduce this to
$$\sum_{k=0}^{3} {n \choose k} (x^3+3x)^{(k)}(\sin x)^{(49-k)}$$
$$= {49 \choose 0} (x^3+3x)^{(0)} (\sin x)^{(49)} + {49 \choose 1} (x^3+3x)^{(1)} (\sin x)^{(48)} + {49 \choose 2} (x^3+3x)^{(2)} (\sin x)^{(47)} + {49 \choose 3} (x^3+3x)^{(3)} (\sin x)^{(46)}$$
Using the derivatives of sines and cosines, this comes down to
$$= {49 \choose 0} (x^3+3x) \cos x + {49 \choose 1} (3x^2+3) \sin x + {49 \choose 2} (6x) (-\cos x) + {49 \choose 3} 6 (-\sin x)$$
I'll leave the number crunching and simplification for you.
A: So the general Leibniz rule somewhat is similar to the binomial theorem. Let $u,v$ be n-th differentiable functions then $(uv)^{(n)}=\sum_{k=0}^{n}{n \choose k}u^{(k)}v^{(n-k)}$ where the exponents denote the $kth$ derivative.
When now note that the fourth derivative and higher of $x^3+3x$ will be zero so we only need to consider the cases where the derivative taken is less than fourth one of $x^3+3x$ . For the $sin(x)$ we use the general formula that $(sin(x))^{(n)}=sin(\frac{n\pi}{2}+x)$. 
Thus the forty-nith derivative will be. 
$((x^3+3x)sin(x))^{(49)}={49 \choose 46}6sin(23\pi+x)+{49 \choose 47}6xsin(\frac{47\pi}{2}+x)+{49 \choose 48}(3x^2+3)sin(24\pi+x)+(x^3+3x)sin(\frac{49\pi}{2}+x)$
Which wil simplfy to 
$(x^3 + 3 x) cos(x) + 49 (3 x^2 + 3) sin(x) - 110544 sin(x) - 7056 x cos(x)$
