Bounding f′ by f and f″ (Landau's inequality)

Taylor–Lagrange. For $h>0$ there exist $c_1\in(x,x+h)$ and $c_2\in(x-h,x)$ with
$$f(x+h)=f(x)+hf'(x)+\tfrac{h^2}2f''(c_1),\qquad f(x-h)=f(x)-hf'(x)+\tfrac{h^2}2f''(c_2).$$

Subtracting.
$$2hf'(x)=f(x+h)-f(x-h)-\tfrac{h^2}2\bigl(f''(c_1)-f''(c_2)\bigr).$$
Using the bounds, $2h|f'(x)|\le2M_0+h^2M_2$, that is
$$|f'(x)|\le\frac{M_0}h+\frac{hM_2}2\qquad\text{for every }h>0.$$

Optimising. The right side is minimal at $h=\sqrt{2M_0/M_2}$, where it equals $\sqrt{\frac{M_0M_2}2}+\sqrt{\frac{M_0M_2}2}=\sqrt{2M_0M_2}$. (If $M_2=0$, let $h\to\infty$: $f'=0$.)

Check. For $f=\sin$: $M_0=M_2=1$ and the bound is $\sqrt2\ge1=\sup|\cos|$ ✓.

Remark. A function cannot be both small and have a small second derivative while having a large first derivative: this is a quantitative version of "a smooth bounded function cannot oscillate quickly with large amplitude". The sharp constant is $\sqrt{2M_0M_2}$.