What is the derivative of $H(f(x),g(x)) = f(x)*g(x) $ with respect to $f(x)$

Question

I've been mulling over this quite a bit and I can't quite seem to figure it out. For context, I was given the equation for the volume of a sphere and asked to find the derivative of the volume with respect to the surface area. Now, in this particular example, we know that a sphere's volume is $\frac{4}{3}*\pi*r^3$ and the surface area is $4*\pi*r^2$.

Therefore we could write this as $\frac{4}{3}*\pi*r^3 = 4*\pi*r^2*\frac{r}{3}$ Which, equivalently, is $Volume \ of \ Sphere = Surface\ Area\ of\ Sphere * \frac{r}{3}$

The book goes on to solve this problem by basically solving for $\frac{dV}{dS}$ using the chain rule...(i.e. by algebraically manipulating $\frac{dV}{dS} *\frac{ dS}{dr} = \frac{dV}{dr}$, which is certainly fine...but seems slightly hand wavy and precludes me from really understanding what is going on).

I realized that $\frac{r}{3}$ is essentially a function of $r$...and therefore thought that maybe by generalizing it I could have a deeper insight into the problem. So, as the title inquires, how does one go about finding the following derivative?

Derivative of $H(f(x),g(x)) = f(x)*g(x) $ with respect to $f(x)$

Answer 1

What's really going on here is there are a lot of hidden compositions with inverse functions. You're initially given volume and surface area as a function of radius. Or more technically, you're given two functions $V:[0,\infty) \to \Bbb{R}$, and $S:[0,\infty) \to \Bbb{R}$ defined by \begin{align} V(r) = \dfrac{4\pi r^3}{3} \quad \text{and} \quad S(r) = 4 \pi r^2 \end{align} Now, note that $S$ is an invertible function with inverse $S^{-1}:[0,\infty) \to[0,\infty) \subset \Bbb{R}$ defined by \begin{equation} S^{-1}(\sigma) = \sqrt{\dfrac{\sigma}{4\pi}} \end{equation}

(i.e a surface area of $\sigma$ corresponds to a radius of $\sqrt{\frac{\sigma}{4\pi}}$) So, when you wrote the symbol $\dfrac{dV}{dS}$, you were thinking of volume as a function of surface area. More technically, you were considering the derivative of the composite function $V \circ S^{-1}$. So, now, let's compute its derivative \begin{align} (V \circ S^{-1})'(\sigma) &= V'(S^{-1}(\sigma)) \cdot (S^{-1})'(\sigma) \tag{directly chain rule} \\ &= V'(S^{-1}(\sigma)) \cdot \dfrac{1}{S'(S^{-1}(\sigma))} \tag{$*$}\\ &= S(S^{-1}(\sigma)) \cdot \dfrac{1}{S'(S^{-1}(\sigma))} \end{align} It is easy to verify that $V'(r)=S(r) = 4 \pi r^2$, that's why in the third line, I changed $V'$ to $S$. Also, if you look up the formula for the derivative of an inverse function, that's where the second term came from. Substituting everything, we get: \begin{align} (V \circ S^{-1})'(\sigma) &= \sigma \cdot \dfrac{1}{8\pi\left(\sqrt{\dfrac{\sigma}{4\pi}} \right)} \\ &= \dfrac{\sqrt{\sigma}}{4 \sqrt{\pi}} \end{align}

This is how a technically precise computation would proceed if you carefully mention where all the derivatives are being evaluated. However, as you can see, it is very cumbersome. Hence, people just shorten it to \begin{align} \dfrac{dV}{dS} = \dfrac{dV/dr}{dS/dr} = \dfrac{4 \pi r^2}{8 \pi r} = \dfrac{r}{2} = \dfrac{\sqrt{S}}{4 \sqrt{\pi}} \end{align} if $S = 4 \pi r^2$, or equivalently, $r = \sqrt{\frac{S}{4 \pi}}$.

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By the way, formula $(*)$ is as general as you can get, and assuming you write it very pedantically. In formula $(*)$, I made no use of the given formula for $V$ and $S$; only the chain rule.

Answer 2

An alternative method from the book's solution is to write $$S(r) = 4\pi r^2, \quad V(r) = \frac{4}{3}\pi r^3,$$ solve for $r$ in terms of $S$: $$r = \sqrt{\frac{S}{4 \pi}},$$ then $$V(S) = \frac{4}{3} \pi \left(\sqrt{\frac{S}{4\pi}}\right)^3 = \frac{ S^{3/2}}{6 \pi^{1/2}}.$$ Then $$\frac{dV}{dS} = \frac{1}{6\pi^{1/2}} \frac{3}{2} S^{1/2} = \frac{S^{1/2}}{4\pi^{1/2}}.$$ Expressed in terms of $r$, this is simply $r/2$, which is consistent with using the formula $$\frac{dV}{dS} \cdot \frac{dS}{dr} = \frac{dV}{dr}.$$

Answer 3

This might not be rigorous but I am going to assume that $f$ and $g$ are continuous and differentiable everywhere. This thus reduces to a problem of applying a "product rule".

Here, $\frac{d}{df}(f*g) = g + f*\frac{dg}{df}$

Thus, for your problem, since we have $V = S \times r/3$, then $\frac{dV}{dS} = \frac{r}{3} + S \times \frac{d(r/3)}{dS}$.

Here, we evaluate $\frac{dS}{d(r/3)} = 3 \frac{dS}{dr} = 3 * \frac{d}{dr}(4 \pi r^2) = 24 \pi r$.

Substitute the expression back in, we thus have $\frac{dV}{dS} = \frac{r}{3} + 4\pi r^2 \times \frac{1}{24\pi r} = \boxed{\frac{1}{2}r}$.