I have just learned about the chain rule but my book doesn't mention the proof. I tried to write a proof myself but can't write it. So can someone please tell me about the proof for the chain rule in elementary terms because I have just started learning calculus.
Calculus – How to Prove the Chain Rule
calculus
Related Solutions
You may use any method that works. A convenient feature of mathematics is that it works no matter which (valid) method you use.
For this kind of problem, it is easiest to ask yourself, "how many 'places' is there an $x$?" By "places", I mean "numerator and denominator". If there are $x$s in both places, then you (probably) will prefer to use the quotient rule. If there are $x$s only in the numerator, then there is no quotient and the result is straightforward. If there are $x$s only in the denominator, then the quotient rule will be very easy; the lack of $x$s in the numerator makes the derivative of the numerator especially simple. Similarly, the chain rule makes it easy in the case that there are only $x$s in the denominator since this is "some function of $x$" raised to "some negative power".
Consider $g(x) = \frac{1}{f(x)}$. We may apply the quotient rule: $$g'(x) = \frac{f(x) \cdot 0 - 1 \cdot f'(x)}{f(x)^2} = \frac{-f'(x)}{f(x)^2}$$ or we may apply the chain rule: $$g'(x) = \left( f(x)^{-1} \right)' = (-1)f(x)^{-2}f'(x) = \frac{-f'(x)}{f(x)^2}$$. We get the same answer either way, so it doesn't matter which method we use.
Some people find it easier to go one way or the other. It doesn't matter which way is chosen as long as the method is applied correctly.
I used to think this, too, until I taught Calculus I.
If you, as a math student and enthusiast, like to see the product rule, etc., as special cases of the multivariate chain rule, then that is good for you and deepens your understanding.
However, my experience has been that reasoning from the general to the specific doesn't always sink in to the novice learner. If the multivariate chain rule is mumbo-jumbo, nothing derived from it is understandable either.
The median student in Calculus I struggles with the concept of function, has trouble working with more than two variables, and can't keep straight whether $\frac{1}{x}$ is the derivative of $\ln x$ or the other way around. I'm not trying to bash Calculus I students; only to recognize that they are in a different place mathematically than we are now, or even than we were when we first learned Calculus I. To reach them, we have to understand where their frontiers are and what is just beyond them.
Best Answer
Assuming everything behaves nicely ($f$ and $g$ can be differentiated, and $g(x)$ is different from $g(a)$ when $x$ and $a$ are close), the derivative of $f(g(x))$ at the point $x = a$ is given by $$ \lim_{x \to a}\frac{f(g(x)) - f(g(a))}{x-a}\\ = \lim_{x\to a}\frac{f(g(x)) - f(g(a))}{g(x) - g(a)}\cdot \frac{g(x) - g(a)}{x-a} $$ where the second line becomes $f'(g(a))\cdot g'(a)$, by definition of derivative.