If $f\left(x\right)=\alpha x^n,$ then $\alpha =$

We have, $f\left(x\right)=\alpha x^n$

Differentiating both sides w. r. t. x, we obtain

$\begin{array}{l} \frac{\mathrm{d}}{\mathrm{dx}}\left\{\mathrm{f}\right(\mathrm{x}\left)\right\}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\alpha {\mathrm{x}}^{\mathrm{n}}\right)\\ \Rightarrow {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\alpha \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{n}}\right)\Rightarrow {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\alpha \mathrm{n}\cdot {\mathrm{x}}^{\mathrm{n}-1}\end{array}$

Putting $x = 1$ on both sides, we get

${\mathrm{f}}^{\mathrm{\prime }}\left(1\right)=\alpha \cdot \mathrm{n}\Rightarrow \alpha =\frac{{\mathrm{f}}^{\mathrm{\prime }}\left(1\right)}{\mathrm{n}}$