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Question: Find the instantaneous rate of change of $x^2 + 16$ with respect to $x^3-1$.

My solution: Let $f(x) = x^2+16$ and $g(x) = x^3 -1$ and $f(g(x))= (x^3-1)^2+16$ . In order to solve problem I have to find $f'(g(x))$.

We know $(f(g(x)))'=f'(g(x))g'(x)$. and $f'(g(x))=$$\frac{(f(g(x)))'}{g'(x)}$. From this we get $f'(g(x))=2(x^3-1)$.

Is this solution true?

Gary
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student0434
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1 Answers1

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There are two approaches to this - both giving the same result.

The first approach is what explains the term "derivative" (or "instantaneous rate of change") of a function "with respect to another function". Let's fix $x\in\mathbb R$. In some neighbourhood of $x$, the function $g(x)$ will be invertible, i.e. you will be able to express $x$ in terms of $t=g(x)=x^3-1$. (In this particular case, this is true on the whole $x$ axis, because $g(x)$ is $1-1$, but in other cases you may need to restrict the domain to possibly a smaller neighbourhood of $x$.) What you are asked to do is express $x=g^{-1}(t)$, then express $f(x)$ as a function of $t=g(x)$ and then calculate the derivative of that function $f(g^{-1}(t))$ (still as a function of $t$), and finally represent this derivative back as a function of $x=g^{-1}(t)$.

You could directly do this manipulation on given functions $f$ and $g$. If $t=g(x)=x^3-1$, then $x=\sqrt[3]{t+1}$ and $f(x)=x^2+16=(\sqrt[3]{t+1})^2+16$. Thus, the required rate of change is $\frac{df}{dt}=2(\sqrt[3]{t+1})\frac{1}{3}(t+1)^{-2/3}=\frac{2}{3}\sqrt[3]{t+1}\frac{1}{(\sqrt[3]{t+1})^2}=\frac{2}{3}x\frac{1}{x^2}=\frac{2}{3x}$.

However, let's apply the rules of calculus without rushing to substitute $f(x)$ and $g(x)$:

$$\frac{d[f(g^{-1})(t)]}{dt}=f'(g^{-1}(t))\frac{d[g^{-1}(t)]}{dt}=f'(g^{-1}(t))\frac{1}{g'(g^{-1}(t))}=\frac{f'(g^{-1}(t))}{g'(g^{-1}(t))}=\frac{f'(x)}{g'(x)}$$

The second approach is unjustified (or, rather, is justified by the first approach!) but is easy to memorize. You may have learned it in this form. The idea is: what you want to find "is" $\frac{df}{dg}$, so let's "extend" this fraction by "dividing" both the numerator and denominator by $dx$, which gives:

$$\frac{df}{dg}=\frac{df/dx}{dg/dx}=\frac{f'(x)}{g'(x)}$$

Now, with whichever approach, you can substitute the actual derivatives of $f(x)$ and $g(x)$, and the result is just $\frac{2x}{3x^2}=\frac{2}{3x}$.

  • @student0434: Yes, but let's not mix $x$ and $t$. In your problem $x$ is used to denote a parameter. You have been, sort of, given a curve $(x^3-1, x^2+16)$ and you want to find the derivative (i.e. slope of the tangent to that curve) - again as a function of $x$. So you would call $t=x^3-1$, express $x^2+16=(\sqrt[3]{t+1})^2+16$, differentiate that on $t$ (getting $2\sqrt[3]{t+1}\frac{1}{3}(t+1)^{-2/3}$) and then replace back $t=x^3-1$ to get $\frac{2}{3}x\frac{1}{x^2}=\frac{2}{3x}$. It is just that, after a long process, you end up with ... $f'(x)/g'(x)$ –  Mar 29 '21 at 10:56