Finding a rate of sphere area increase given its volume increaseVolume of sphere, V =...

Finding a rate of sphere area increase given its volume increase
Volume of sphere, $V={\displaystyle \frac{4}{3}}\pi r^3$
Surface area of sphere $S=4\pi r^2$.
If we know, ${\displaystyle \frac{dV}{dt}}=R$.
Let us consider both volume and area as composite functions, thus ${\displaystyle \frac{dV}{dt}}={\displaystyle \frac{dV}{dr}}\times {\displaystyle \frac{dr}{dt}}=4\pi r^2\times {\displaystyle \frac{dr}{dt}}=R$
whence ${\displaystyle \frac{dr}{dt}}={\displaystyle \frac{R}{4\pi r^2}}$ since ${\displaystyle \frac{dS}{dt}}=6\pi r{\displaystyle \frac{dr}{dt}}$, let the value of the ${\displaystyle \frac{dr}{dt}}$ into the second equation, to get the answer. Is this approach logically correct?