# Basic differentiation question on derivative of conical volume

So I was reading Polya's book and in it, there was a problem involving finding the rate of change of depth of water in a cone.

At some point, we come to the conclusion that V = $\pi a^2 y^3/(3b^2)$ where $V$ = volume of water, $a$= radius of base of cone, $y$ = depth of water and $b$ = height of cone.

This is where I get lost...They then come to the conclusion that since $V$ increases along with $y$, then upon differentiation, the result should be $dV/dt = (\pi a^2 y^2 / b^2 )dy/dt$.

How come they're multiplying with $dy/dt$? whats the rule ? I know this must be extremely basic but it's been a while since I touched maths and wanted to get back into it.

• Chain rule: $\frac{dV}{dt} = \frac{dV}{dy}\frac{dy}{dt}$ – Ben Grossmann Aug 6 '15 at 15:54
• $y$ is a function of $t$, so they apply the chain rule. – BadAtMaths Aug 6 '15 at 15:58

The answer is easy: $$\frac{dV}{dt} = \frac{d}{dt} \frac{\pi a^2}{3b^2} y^3 = \frac{\pi a^2}{3b^2} \cdot 3 y^2 \frac{dy}{dt}$$ by the chain rule applied to $t \mapsto y(t)^3$.
• So...would it be safe to say that d/dt $x^4$ = 4$x^3$dx/dt ? – Newbie Aug 6 '15 at 17:41
• Yes, it is even correct when $x$ is independent of $t$. – Siminore Aug 6 '15 at 18:44