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Chain Rule - Supporting Problem 3

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Lesson Parent:

Differentiation IV: Differentiation Formulas Everyone Must Know

Problem:

Prove that if

$g(x)$ is differentiable at $a$
for all $\bar{\delta}\gt0$ there exists an $\bar{x}$ in the interval $(a-\bar{\delta},a)\cup(a,a+\delta)$ such that $g(x)=g(a)$

are true then $g'(a)=0$ .

Answer:

$g'(a)=0$

Solution:

This problem is needed to help prove the Chain Rule.

The intuition here is that as <Sign in to see all the formulas> gets smaller and smaller then $g(x)$ is basically a constant value of $g(a)$ . Because the derivative of a constant is zero it seems that $g'(a)$ should be zero here. Now let's prove it.

Note that this is the form of the derivative definition we are using.

By the assumption of the problem $g'(a)$ exists which means that for every <Sign in to see all the formulas> there exists a <Sign in to see all the formulas> such that if <Sign in to see all the formulas> then

<Sign in to see all the formulas>

Furthermore, by the assumption of the problem, in every interval <Sign in to see all the formulas> there is an <Sign in to see all the formulas> where <Sign in to see all the formulas>. Since <Sign in to see all the formulas> in taking the limit, <Sign in to see all the formulas> making

<Sign in to see all the formulas>

Therefore

<Sign in to see all the formulas>

for at least one value of

$x$ in every interval <Sign in to see all the formulas>.

Now since this is done for every value of <Sign in to see all the formulas>, we have a <Sign in to see all the formulas> and an $x$ in <Sign in to see all the formulas> where

<Sign in to see all the formulas>

The only way this can be true for every <Sign in to see all the formulas> is if <Sign in to see all the formulas>. Otherwise there would be some <Sign in to see all the formulas> and <Sign in to see all the formulas> where

<Sign in to see all the formulas>

if <Sign in to see all the formulas>.

The proof is now complete.