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.

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.
