Using chain rule to differentiate $f(x)=a(x)b(x)$?

Question

Why can I not apply the chain rule to a product in the following way.

If we have some product:

$$f(x)=a(x)b(x)$$

Consider the multiplication of b by a as another’s function so that:

$$f(b(x))=ab$$

So that

$\frac{df}{dx} = f’(b)b’(x)$

Something feels very wrong. But I can’t put my finger on it.

Answer 1

If $f(x)=a(x)x$ (it looks as if this is what you have in mind), then $f\bigl(b(x)\bigr)$ is equal to $a\bigl(b(x)\bigr)b(x)$, instead of $a(x)b(x)$.

Answer 2

You can't use the chain rule because this is not a composition. Just calling it one does not make it one. For example, suppose $a(x)=x+1$ and $b(x)=x^2$. Then your first line says $$ f(x) = x^2(x+1) = x^3+ x^2 $$ so $$ f(b(x)) = f(x^2) = (x^2)^3 + (x^2)^2 $$ so not the same as "$ab$".

Answer 3

Write $\mu(u,v)=uv$. Then $$ \mu_1(u,v) = \frac{\partial\mu}{\partial u}(u,v)= v, \quad \mu_2(u,v) = \frac{\partial\mu}{\partial v}(u,v)= u $$

Now $f(x)=\mu(a(x),b(x))$ and so $$f'(x)=\mu_1(a(x),b(x))a'(x)+\mu_2(a(x),b(x))b'(x)=b(x)a'(x)+a(x)b'(x)$$

This is the multiplication rule using the chain rule.

Answer 4

Based on the remark given by David Mitra in the comments, you are considering: $$ g(b(x))=a(x)b(x) $$ which implies we must have $$ g(x)=a(b^{-1}(x))x $$ Finding the derivative of this $g$ will be a bit messy. $$ \frac d{dx}g(x)=a'(b^{-1}(x))\cdot\frac 1{b'(b^{-1}(x))}\cdot x+a(b^{-1}(x)) $$ and so plugging in $b(x)$ and applying the chain rule gives us: $$ \begin{align} \frac{d}{dx}g(b(x))&=\left(a'(x)\cdot\frac 1{b'(x)}\cdot b(x)+a(x)\right)\cdot b'(x)\\ &=a'(x)b(x)+a(x)b'(x) \end{align} $$

Answer 5

Also re-interpreting as per David Mitra's comment, that you are looking for some function $g$ such that $g(b(x)) = a(x)b(x)$ (where $g$ is something different from $f(x) = a(x)b(x)$).

Your problem here is that then $a(x) = \frac {g(b(x))}{b(x)}$. I.e., the value of $a(x)$ depends on that of $b(x)$. What happens when $b(x_1) = b(x_2)$ but $a(x_1) \ne a(x_2)$? For example, suppose $b(x) = c$ for some constant $c$, but $a(x) = x$? Clearly no such $g$ can exist.

So you cannot turn every product into a composition of single variable functions. However, as lhf has pointed out, multiplication itself is a multivariate function, and the product rule is actually just an application of the multivariate chain rule.

Answer 6

Alternatively, if we know the chain rule and the derivative of the logarithm as well as the fact that $\log(ab)=\log(a)+\log(b)$, we get that \begin{align} \left(\log(f(x)g(x)\right) &= \left(\log(f(x))+\log(g(x))\right)'\\ &= \frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)} \end{align} But we also have that \begin{align} \left(\log(f(x)g(x)\right) &= \frac{\left(f(x)g(x)\right)'}{f(x)g(x)} \end{align}

Answer 7

If you can find such a composition $A\circ b = a(x)b(x)$, then the chain rule gives you the product rule. You can find $A\circ b$ just if b is invertible.

• Because $a(x)$ and $b(x)$ are both functions of $x$, you might not be able to find a "multiplication function" $A$ that lets you write the product $a(x)b(x)$ as a chain rule composition $A(b(x))$.

  This is because $A$ only sees the output of the function $b(x)$. It doesn't see what $x$ is. If it can't see what $x$ is (or figure it out based on the value of $b(x)$), then it can't compute $a(x)$ and so it can't compute $a(x)b(x)$.

• When you can recover the value of $x$ from the value of $b(x)$, we say that $b$ is invertible.

• If $b$ is invertible, then it has an inverse $b^{-1}$ which recovers the value of $x$. So, if $y=b(x)$ then $b^{-1}(y) = x$.

  If $b$ is invertible, then you can define the function $A(y) \equiv a(b^{-1}(y)) \cdot y$.

• Then notice that by definition, $A(b(x)) = a(b^{-1}(b(x))) \cdot b(x) = a(x)b(x)$, exactly as we want it. We have written $a(x)b(x)$ as a chain rule composition $A(b(x))$.

• When we take the derivative using the chain rule, we find that $\frac{d}{dx}A(b(x)) = a^\prime(x)b(x) + a(x)b^\prime(x)$, exactly as you want. Though because the definition of $A$ depends on a product, you either have to use the product rule anyways to differentiate it, or use the limit definition of the derivative to compute it.

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In multivariable calculus, the chain rule directly gives you the product rule.

What you might be looking for, after all, is a way to use the chain rule as a fundamental tool for proving all of the other rules such as the product rule. You can do this with multivariable calculus.

• The multiplication function is $M(u,v) = uv$. Its derivative is $DM(u,v) = \langle v, u\rangle$.

• If your two functions are $a(x)$ and $b(x)$, then their product is $g(x) = a(x)b(x)$.

• We can write $g$ as a composition: $g(x) = M(a(x), b(x))$.

• The derivative of $g$ is therefore the derivative of $M(a(x), b(x))$---and we can compute this latter derivative using the chain rule.

• By the chain rule,

\begin{eqnarray*} D\left[M(a(x),b(x))\right] &= DM( a(x), b(x))\bullet D\langle a(x), b(x)\rangle\\ &= \langle b(x), a(x)\rangle \bullet \langle a^\prime(x), b^\prime(x)\rangle\\ &= b(x)a^\prime(x) + a(x)b^\prime(x) \end{eqnarray*}