Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm following along on this proof of the chain rule. All is clear to me except the step where they say:

Differentiablility implies continuity; therefore $\Delta_u \to 0$ as $\Delta_x \to 0$ in $f\big(g(x)\big)$.

Could someone explain to a calculus newb why that's the case, in simple terms?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

A piece of advice: try another site as the one you linked uses what I consider pretty poor notation.

If we have $\,f(g(x))\,$ then I'd go

$$\lim_{x\to x_0}\frac{f(g(x))-f(g(x_0))}{x-x_0}=\lim_{x\to x_o}\frac{f(g(x))-f(g(x_0))}{g(x)-g(x_0)}\frac{g(x)-g(x_0)}{x-x_0}$$

And what that site says with its cumbersome (for me) notation is that $\,x\to x_0\Longrightarrow g(x)\to g(x_0)\,$ , since differentiability implies continuity.

Some care must be put to avoid the possibility of dividing by zero in the above, of course.

share|improve this answer

(before answering your question, let me tell you that I don't really recommend that proof; the main problem is that derivatives are calculated at a point, and the notation used in the proof obscures that fact completely)

You are calculating your derivatives at some $x=x_0$. Then $\Delta x=x-x_0$; and $\Delta u=u(x)-u(x_0)$. As $u$ is differentiable, it is continuous. Continuity at $x_0$ for $u$ means exactly that $u(x)-u(x_0)\to0$ when $x-x_0\to0$, i.e. $\Delta u\to 0$ when $\Delta x\to 0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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