What are the implications of x=0 in terms of function derivability?

[Noismaker]

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Thank you

 

[HallsofIvy]

I think "differentiability" would be a better word than "derivability" since that might be confused with the ability to "derive" a formula. In any case, the first thing I would do is remove the "absolute value" by looking at x positive or negative separately.

If x is positive, then $$f(x)= x^x$$ and, taking the logarithm of both sides, log(f(x))= x log(x). Differentiating, $$f'(x)/f(x)= log(x)+ 1$$ so that $$f'(x)= (log(x)+ 1)f(x)= (log(x)+ 1)x^x$$. That exists for all positive x so f(x) is differentiable for all positive x.

If x is negative, make the substitution y= -x. |x|= -x= y so that $$x^x= y^{-y}$$. Again take the logarithm of both sides of $$f(y)= y^{-y}$$, $$log(f(y))= -ylog(y)$$. Differentiating, $$f'(y)/f(y)= -log(y)- 1$$ so that [tex]f'(y)= -f(y)(log(y)+ 1)= -y^y(log(y)+ 1) and, since dy/dx= 0, $$f'(x)= (-x)^x(log(|x|)+ 1)$$. That exists for all negative x so f(x) is differentiable for all negative x.<br /> <br /> Finally, look at x= 0. The derivative of f(x) at x= 0 is $$\lim_{h\to 0} \frac{f(h)- f(0)}{h}= \lim_{h\to 0}\frac{h^h- 1}{h}$$. Does that limit exist?[/tex]

 

[Noismaker]

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so in x=0 we have a point of non-differentiation, cusp, angular point, or flattened to vertical tangent.