the derivative of $f(x)=|x|^{\frac 52}$ From my point of view answer must be $\frac 52|x|^{\frac 32}$...but answer in my text book is $\frac 52 x |x|^{\frac 12}$....help me to solve it.I have tried it to break the function for positive and negative parts and directly differentiated it but the answer is not matched
 A: For $x \geq 0, $ function is $x^{\frac{5}{2}}$; for $x < 0$, it is $(-x)^{\frac{5}{2}}$.
Differentiation for $x > 0$ is $\frac{5}{2}x^{\frac{3}{2}}$;
Differentiation for $x < 0$ is $\frac{5}{2}(-x)^{\frac{3}{2}}(-1) = -\frac{5}{2}(-x)^{\frac{3}{2}}$ which is same as $\frac{5}{2}x|x|^{\frac{1}{2}}$ when $x < 0$;
As pointed out by GEdgar, we must check the derivative at $x = 0$ separately. In this case, note that $x^{\frac{5}{2}} \to 0$ as $x \to 0^+$ and the derivative $\frac{5}{2}x^{\frac{3}{2}} \to 0 $ as $x \to 0^+$. 
So we can say that the even extension $(-x)^\frac{5}{2}$ makes the complete function differentiable with the value of derivative equal to $0$ at $x = 0$.
A: You can apply chain rule using, $\frac{d|x|}{dx}=\frac{x}{|x|}$
$$f(x)=|x|^{\frac{5}{2}}$$
$$f'(x)=\frac{5}{2}|x|^{\frac{3}{2}}\frac{x}{|x|}$$
$$f'(x)=\frac{5}{2}x|x|^{\frac{1}{2}}$$
A: Breaking it up into a piecewise function is fine.
$f(x) = x^{\frac 52}$ (for $x \geq 0$) [$1$]
$f(x) = (-x)^{\frac 52}$ (for $x < 0$) [$2$]
Differentiating them separately,
$f'(x) = \frac 52x^{\frac 32}$(for $x \geq 0$) [$1$]
$f'(x) = -\frac 52(-x)^{\frac 32}$(for $x < 0$) [$2$]
The second part can be rearranged to:
$f'(x) = -\frac 52(-x)(-x)^{\frac 12} = \frac 52 x (-x)^{\frac 12} = \frac 52 x|x|^{\frac 12}$, which holds for $x < 0$. Note where the minus signs cancel out.
And for the first piece, a similar rearrangement holds trivially, 
$f'(x) = \frac 52x^{\frac 32} = \frac 52 x (x)^{\frac 12} = \frac 52 x|x|^{\frac 12}$, which holds for $x \geq 0$.
So $f'(x) = \frac 52 x|x|^{\frac 12}$ for the entire domain.
Your book is right.
A: $f(x)=|x|^\frac 5 2$.
Using power rule and chain rule:
$f'(x)=\frac 5 2  |x|^\frac 3 2 * {d\over dx} |x|)$. Now we have to find ${d\over dx} |x|$. $|x|$ equals $-x$ if $x<0$, and $|x|=x$ if $x\ge0$, from here it is easy to see that the derivative of $|x|$ is uqual to ${x \over |x|}$. We put this in the expression above and we get:
$f'(x)=\frac 5 2  |x|^\frac 3 2 * {x \over |x|}=\frac 5 2  |x|^\frac 1 2 * x$. Note that $x$ can not equal $0$, because the derivative of $|x|$ is undefined at $x=0$.
A: Remark for the point $x=0$.  
First another exmaple:
$$
g(x) = \begin{cases}
|x|^{5/2} + 1, \quad x\ge 0
\\
|x|^{5/2}+2, \quad x< 0
\end{cases}
$$
If the technique you used for $f(x)$ also works for $g(x)$, there must be something missing for $x=0$.  
Certainly the existing proofs all work to show $g'(x) = \frac 52 x |x|^{\frac 12}$ for $x \ne 0$.  But in this case $g'(0)$ does not exist.  
Therefore, in the case of $f(x)$, something more is needed.
For example apply the definition ...
$$
f'(0) = \lim_{h \to 0}\frac{f(0+h) - f(0)}{h} = \lim_{h \to 0}\frac{|h|^{5/2}}{h}
$$
if the limit exists.  Then compute that the left and right limits are both $0$, therefore the limit itself is $0$.
