0
$\begingroup$

This question is about understanding the implications of the chain rule, and I will use the area of a triangle to demonstrate:

$$A=\frac{B\cdot H}{2}$$

Let's say that it is growing (both the base and the height and thus the area too) and we want to find the rate of change of one of its components. So we differentiate everything with respect to time:

$$A'=\frac{1}{2} (BH)'$$

I know that I should use that product rule for $(BH)'$, but I was curious about using the power rule. Consider that $(BH) = (BH)^1$, so by the power rule we have $(BH)' = 1$, and then by the chain rule we multiply by $(BH)'$ again:

$$A'=\frac{1}{2} \cdot 1\cdot(BH)'$$

Obviously that doesn't help. BUT, it made me think: whenever we differentiate something, the chain rule will always trigger, right? It just happens that depending on the differentiation method you use, it may be pointless like just now. So let's use the product rule now:

$$A'=\frac{1}{2} \cdot 1\cdot(B'H + H'B)$$

I used the product rule to differentiate $BH$. But, by the chain rule, should I not be multiplying by the derivative of $(BH)$? In other words:

$$A'=\frac{1}{2} \cdot 1\cdot(B'H + H'B)\cdot (BH)'$$

And if I keep going with the product rules, I would have

$$A'=\frac{1}{2} \cdot 1\cdot(B'H + H'B)\cdot (B'H + H'B) \cdot (B'H + H'B) \cdot (B'H + H'B) \cdot ...$$

Because of the chain rule!

Again, this is based on my understanding that the chain rule happens whenever we differentiate, and the product rule is a form of differentiation. This example is clearly wrong, but I don't know why is that the case.

What is wrong with my interpretation of the chain rule?

$\endgroup$
4
$\begingroup$

The paradox is explained by noticing that you're using the notation $(BH)'$ to mean two different things.

First, we're assuming (I suppose) that $B$ and $H$ are functions of another variable, such as time $t$; and then the first $(BH)'$ means $\frac d{dt}(BH)$.

After you apply the chain rule, writing $A' = \frac1\cdot1\cdot(B'H+H'B)\cdot(BH)'$, the context in which that calculation makes sense is if $(BH)'$ is interpreted as $\frac d{d(BH)}(BH)$. And that expression simply equals $1$. (You're implicitly writing $BH = f(g(B,H))$ where $g(B,H)=BH$ and $f(BH)=BH$, that is, $f(x)=x$, whose derivative is $f'(x)=1$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.