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Here is a really basic problem. Let $R$ be the radius of a sphere, and draw this uniformly on $[1,10]$ so that $R$ has density $$f_R(x)=\frac{1_{1<x<10}}{9}.$$ Let $V=g(R)=\frac{4\pi}{3}R^3$ be the volume of this sphere.

We want to compute the density of $V$. The usual way is: $R=g^{-1}(V)=\left(\frac{3V}{4\pi}\right)^{1/3}$, so $$\begin{aligned} f_V(y) &=f_R(g^{-1}(y))\left\lvert\tfrac{d}{dy}g^{-1}(y)\right\rvert\\ &=\frac{1_{\frac{4\pi}{3}<y<\frac{4\pi}{3}1000}}{9} \frac13\left(\frac3{4\pi}\right)^{1/3}y^{-2/3}\\ &=1_{\frac{4\pi}{3}<y<\frac{4\pi}{3}1000} y^{-2/3} \frac1{27}\left(\frac3{4\pi}\right)^{1/3}. \end{aligned}$$

Then I thought, hey, let's try Leibniz notation for once. So, dropping the indicators, I wrote $$ \frac{dR}{dx}=\frac19 \qquad\text{and}\qquad V=\frac43\pi R^3,$$ so clearly $$\frac{dV}{dR}= 4\pi R^2=4\pi\left(\frac{3V}{4\pi}\right)^{2/3},$$ and $$ \frac{dV}{dx}=\frac{dV}{dR}\frac{dR}{dx}=4\pi\left(\frac{3V}{4\pi}\right)^{2/3}\cdot\frac19, $$ but of course this is dead wrong.

I'm curious, where have I gone wrong in my Leibniz notation? I never use it and I thought this was the way to go. I'm suspicious of my $\frac{dV}{dR}$, maybe the problem is there.

mostsquares
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1 Answers1

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So you have $F_V(V)=F_R(g^{-1}(V))$, and so $F_V'(V)=f_V(V)=F_R'(g^{-1}(V)) (g^{-1})'(V)=f_R(g^{-1}(V)) (g^{-1})'(V)$. That's the chain rule. In Leibniz notation this reads $\frac{dF}{dV}=\frac{dF}{dR} \frac{dR}{dV}$.

The rule for inverse functions is $(g^{-1})'(V)=\frac{1}{g'(R(V))}$ where $g(R(V))=V$. In Leibniz notation that reads $\frac{dR}{dV} = \frac{1}{\frac{dV}{dR}}$, which is a pretty intuitive thing.

So then you are left to evaluate $\frac{dR}{dV}$, which you can do either by solving for $R(V)$ and differentiating, or by differentiating $V$ first, taking the reciprocal and then plugging $R(V)$ into it.

Ian
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