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Using the definition of the derivative, you are required to prove that

$$g(x) = \sqrt{\frac{x^2(1+x)}{1-x}}, -1\le x \lt1$$

is NOT differentiable at x = 0. I know there exists a cusp at x = 0, but I need to use the definition of the derivative. Also I know that if g'(0) exists, then g(x) is differentiable at x = 0. Is it then okay to say that if g'(0) does not exist, then g(x) is not differentiable at x = 0?

chris24
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1 Answers1

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Yes, of course: if the limit of the Newton ratio doesn't exist, the function is not differentiable.

Hint: $$ g(x)=|x|\sqrt{\frac{1+x}{1-x}} $$ so you want to compute $$ \lim_{x\to0^+}\frac{g(x)-g(0)}{x}= \lim_{x\to0^+}\frac{|x|}{x}\sqrt{\frac{1+x}{1-x}} $$ What about the limit for $x\to0^-$?

egreg
  • 238,574