How to find dy/dx when function defined at certain number Let function y=(sin x)^x and defined at x in the element of (0,pi).
Find dy/dx.
I get the answer for dy/dx =sin^x (x) (In (sin(x))+ x cot (x))
But when I insert 0 or pi into this, calculator shows undefined.
What should I do this question?
 A: The domain of this function is $(0,\infty)$, so $x=0$ is not even in the domain. And since $x=0$ is not in the domain, the derivative is not defined at $x=0$.
And that answers your question. You can stop now.

Or, you could push this problem a lot further, if you so desired and were sufficiently curious and energetic.
For starters, you could ask yourself: Why was $x=0$ left out of the domain? Can't I just plug $x=0$ into that formula?
Well, if you do that then you'll get $0^0$ which is a well-known indeterminate form: the exponentiation $0^0$ does not have a well-defined value.
But then you could push this problem still further: Could I assign a value at $x=0$?
Well, if you want to do that then the value you assign should satisfy a strong constraint: the function must be continuous at $x=0$, because if it has a derivative at $x=0$ then it must be continuous at $x=0$; so the only choice for the value of the function at $x=0$ is
$$\lim_{x \to 0^+} (\sin x)^x
$$
But now a new problem raises its ugly head: Does that limit even exist?
As it turns out, that limit does exist:
$$\lim_{x \to 0^+} (\sin x)^x = 1
$$
The proof of this is tricky, it uses the method of "logarithmic l'Hopital's rule: first one computes the limit of $\ln(\sin(x)^x) = x \ln(\sin(x))$ using l'Hopital's rule
\begin{align*}
\lim_{x \to 0^+} x \ln(\sin(x)) 
&= \lim_{x \to 0^+} \frac{\ln(\sin x)}{1/x} \quad\bigl(\text{which gives $\frac{-\infty}{\infty}$, so we can apply l'Hopital}\bigr)
\\
&= \lim_{x \to 0^+} \frac{\cos(x)/\sin(x)}{-1/x^2} \\
&= - \lim_{x \to 0^+} \frac{x^2 \cos(x)}{\sin(x)} \\
&= - \lim_{x \to 0^+} \frac{x}{\sin(x)} \cdot x \cdot \cos(x) \\
&= - \lim_{x \to 0^+} \frac{x}{\sin(x)} \cdot \lim_{x \to 0^+} x = \lim_{x \to 0^+} \cos(x) \\
&= 1 \cdot 0 \cdot 1 \\ 
&= 0
\end{align*}
and then one computes
$$\lim_{x \to 0^+} (\sin(x))^x = e^{\lim_{x \to 0^+} x \ln(\sin(x))} = e^0 = 1
$$
So we can extend the definition to obtain a function which is continuous at $x=0$:
$$f(x) = \begin{cases}
(\sin x)^x & \quad\text{if $x > 0$} \\
1  &\quad\text{if $x=0$}
\end{cases}
$$
At last we have gone far enough that we can re-ask the title question:

Does $f'(0)$ exist, and if so how do we find it?

To answer this question, the only hope is to go straight to the definition of the (one-sided) derivative:
\begin{align*}
f'(0) &= \lim_{x \to 0^+} \frac{f(x)-f(0)}{x-0} \\
      &= \lim_{x \to 0^+} \frac{(\sin x)^x - 1}{x} \\
\end{align*}
But now my story comes to a somewhat abrupt end. L'Hopital's rule fails on this one, because the limit that you get for L'Hopital's rule is the exact same limit you started with.
But when you start plugging in values of $x$ approaching $0$, you can see something strange and weird happening. I plugged in
$$.00000000000001
$$
using Google calculator and got $-32.2408766351$. I suspect that the limit is $-\infty$ and so $f(x)$ is not differentiable at $x=0$.
One can poke around and find a few other hints that this might be true. For example, use the approximation $\sin(x) \approx x$ which is good to second order, then type in "graph of $x^x$" into Google and zoom in to the point $(0,1)$. It sure looks like it has a vertical tangent!
A: The way I read what you've written, you aren't asked to find the derivative at $x=0$ or $x=\pi$. You're asked to find the derivative everywhere in-between those two points, but not at those two points. So there is no issue here.
