How do you know when to use the Chain Rule instead of the Power Rule?

Question

We use the chain rule for functions that can be seen as composed functions. However, for a simple function like $y=x^2$ we can also say that this is just a composition of 2 functions: $y=x$ and $y=(\text{something})^2$

Now, if we use the chain rule instead of the power rule, it still works and we still get $y'=2x$ But I'm wondering if the "power rule" is really just a shortcut of the chain rule for polynomials (polynomials meaning, no composite functions).

*NOTE: the answer was given to me in one of comments, which is basically that the power rule is applied when the composed functions are only the identity function (f(x)=x) and that of course applies only in polynomials.

Answer 1

I would say that you can always make a power rule problem into a chain rule problem in an unenlightening way. Yes, if $f(x) = x^n$, then we can define $g(x) = x$, so that $$ \frac{d}{dx}f(x) = \frac{d}{dx}f(g(x)) = f'(g(x))g'(x) = n [g(x)]^{n-1} \cdot 1 = nx^{n-1} $$ Note, however, that we would need to know the power rule in order to find the "outer" derivative $f'(x)$.

Slightly more interesting is the fact that if $f(x) = x^n$ and $g(x) = x^m$, then we can write $$ \frac{d}{dx}x^{mn} = \frac{d}{dx}(x^m)^n = \frac{d}{dx}f(g(x)) = n[g(x)]^{n-1} g'(x) = n[x^{m}]^{n-1} \cdot mx^{m-1} =\\ (mn)x^{m(n-1) + m-1} = (mn)x^{mn - 1} $$ Of course, it is easier to use the power rule outright.

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We could derive the power rule (for real numbers $n$) using the chain rule together with the rule $\frac{d}{dx} \ln(x)= \frac 1x$. In particular, if we define $y = x^n$, we can use "logarithmic differentiation". $$ y = x^n \implies\\ \ln(y) = \ln(x^n) \implies\\ \ln(y) = n\ln(x) \implies\\ \frac{d}{dx}\ln(y) = \frac{d}{dx}n \ln(x) \implies\\ \frac{1}y \frac{dy}{dx} = \frac{n}{x} \implies\\ \frac{dy}{dx} = \frac{ny}{x} $$ If we substitute $y$ back in, we find $$ \frac{dy}{dx} = \frac{n(x^n)}{x} = nx^{n-1} $$

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We could also derive the power rule (for integers $n$) using just the multivariate chain rule. In particular, define $$ f(x_1,\dots,x_n) = \prod_{k=1}^n x_n, \quad g(t) = (t,\dots,t) $$ We then find that $$ \frac{\partial f}{\partial x_k} = \frac{1}{x_k} \prod_{k=1}^n x_n, \qquad g'(t) = (1,\dots,1) $$ It therefore follows that $$ \frac{d}{dt}f(g(t)) = \sum_{k=1}^n \frac{\partial f}{\partial x_k}(g(t)) \cdot g'(t) = \cdots = nt^{n-1} $$ so that's kind of neat.

Answer 2

$x^n := \exp (\log x \cdot n)$ hence the derivative is

$\frac nx \cdot\exp (\log x \cdot n) = n \frac{\exp (\log x \cdot n)}{\exp \log x} = n\exp(\log x \cdot n - \log x) = n \exp(\log x(n-1)) = nx^{n-1} $

So yes, with this definition, the power rule is just a special case of the chain rule. This definition of $x^n$ is nice when $n \in \mathbb{R} \setminus \mathbb{Q}$.

Answer 3

Your question is a nonsense, the chain rule is no substitute for the power rule.

Indeed, by the chain rule where you see the function as the composition of the identity ($f(x)=x$) and a power we have

$$(f^r(x))'=f'(x)\frac{df^r(x)}{df}=1\cdot rf(x)^{r-1}=rx^{r-1}.$$

and in this development we needed the power rule anyway.

Answer 4

The chain rule is best defined as: $$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$$ This shows off what the chain rule does mechanically; it allows you to write any part of the function as $u$ and take the derivative in terms of that. If you are simply replacing $x$ with $u$ then the chain rule is applied as: $$\frac{dy}{dx} = \frac{dy}{dx}\frac{dx}{dx}$$ $$\frac{dy}{dx} = \frac{dy}{dx}$$ Which, as you can see above, becomes redundant. As I said above, the chain rule allows you to replace a function with another, and replacing a function ($x$) with the same function ($u$) means you're not actually chain ruling anything.