The maximum slope of curve $\mathrm{y}=-{\mathrm{x}}^3+3{\mathrm{x}}^2+9\mathrm{x}-27\text{ is ......................}$

$\text{ We have, }\mathrm{y}=-{\mathrm{x}}^3+3{\mathrm{x}}^2+9\mathrm{x}-27$
$\text{∴}\text{slope of the curve (}\frac{\mathrm{dy}}{\mathrm{dx}}\text{)}=-3{\mathrm{x}}^2+6\mathrm{x}+9$

$\text{ Now, }\frac{\mathrm{dy}}{\mathrm{dx}}\text{ is consider our main function. }$

$$\begin{gathered} let f\left(x\right)=-3x^2+6x+9 \\ diff w.r.to x on both sides \\ f\text{'}\left(x\right)=-6x+6 \\ and f\text{'}\text{'}\left(x\right)=-6<0 \\ for max or min f\text{'}\left(x\right)=0 \\ \Rightarrow -6x+6=0 \\ \Rightarrow x=1 \end{gathered}$$
So, the maximum slope of given curve is at x=1
The maximum value of the slope is 
${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(\mathrm{x}=1)}=-3\cdot 1^2+6\cdot 1+9=12$