y=|x|(tanx)^(1/2)
find y'
Run the online derivative calculator:
webMathematica Explorations: Step-by-Step Derivatives
Derivative of: x*Tan[x]^(1/2)
With respect to: x
Gives:
$\displaystyle \sqrt{\tan(x)} + \frac{x \sec^2(x)}{2\sqrt{\tan(x)}}$
This is valid for $\displaystyle x > 0$
For $\displaystyle x < 0$, you have
Derivative of: -x*Tan[x]^(1/2)
With respect to: x
which gives:
$\displaystyle \sqrt{\tan(x)} - \frac{x \sec^2(x)}{2\sqrt{\tan(x)}}$
So we can combine these two into one expression for $\displaystyle x \ne 0$:
$\displaystyle \sqrt{\tan(x)} + \left|\frac{x \sec^2(x)}{2\sqrt{\tan(x)}}\right|$
Well, if $\displaystyle y$ is a function of $\displaystyle x$, in other words, $\displaystyle y = f(x)$, you can't do that directly. In this case,
$\displaystyle y(x)=|x|\sqrt{\tan x}$
So we compute the derivative as before to find $\displaystyle y'(x) = \frac{dy}{dx}(x)$. Only once you've done that can you put in $\displaystyle x = 0$ to find what the value of $\displaystyle y'(0)$ is.
But because you have an absolute value, at $\displaystyle x = 0$ the derivative doesn't exist. That's because the slope of the tangent line of $\displaystyle |x|$ changes sign at $\displaystyle x=0$.
We just say it isn't differentiable at x=0 and exclude the point x=0 from the domain of the derivative. You don't need to bother doing anything else.
The domains of the function and its derivative are a bit complicated though (since $\displaystyle \sqrt{\tan x}$ only exists where $\displaystyle \tan x \ge 0$). Are you asked to state the domain of the derivative?
Actually, you are correct! That is the concept of a "weak derivative." This is a bit advanced though. You wouldn't have to see that stuff until you became a graduate student or a senior working in mathematics.
Don't let the following links blow your mind - just be aware that what you are talking about does exist.
Weak derivative - Wikipedia, the free encyclopedia
Subderivative - Wikipedia, the free encyclopedia