Answer
\[
\begin{aligned}
\frac{2 V}{r}=\frac{\partial V}{\partial r} &
\end{aligned}
\]
Work Step by Step
We know that the circular cone of radius $ r $ and the height of $ h $ is
\[
\frac{1}{3} \pi r^{2} h=V
\]
Thus, we we have to evaluate the rate of change of Volume with respect to radius $r$ if $h$ remains constant and prove that
\[
\begin{aligned}
\frac{2 V}{r}=\frac{\partial V}{\partial r} \\
&\frac{\partial}{\partial r}\left(\frac{1}{3} \pi r^{2} h\right)=\frac{\partial V}{\partial r} \\
&\frac{\pi h}{3} \frac{d}{d r}\left(r^{2}\right) =\frac{\partial V}{\partial r} \\
&\frac{2 \pi r h}{3}= \frac{\partial V}{\partial r} \\
&\frac{2}{r} * \frac{1}{3} \pi r^{2} h =\frac{\partial V}{\partial r} \\
\frac{2 V}{r}= \frac{\partial V}{\partial r} &
\end{aligned}
\]
