Why differentiating a function is similar to differentiating its log function? $f(6;p)=\binom{25}{6}p^6(1-p)^6,\quad 0\le p\le1$ 
I have three questions:
$(1)$To find the relative maxima of p, the process is to take the derivative of the function with respect to $p$ equal to $0$ and solving the resulting equation for $p$. 

For easy computation we take the log of the function & differentiate the log function with respect to $p$ equal to $0$ and solving the resulting equation for $p$. 

$\bullet$"My Question Is Why $\frac{d}{dp}f(6;p)$ is similar to $\frac{d}{dp}\log f(6;p)$"
(2) After the step(1) i found that $p=\frac{6}{25}$ is the only root.

But on the book, which i am reading, is written that $p=0,1,\frac{6}{25}$  are the roots.

$\bullet$"How p=0,1 can be root? The denominator $p(1-p)$ becomes $0$ when i do cross multiplication."
(3)To check which value of $p$ maximize $f(6;p)$ , we take the second derivative of the function with respect to $p$ and set $p=0,1,\frac{6}{25}$ respectively. If the second derivative is less than $0$, then $p=$ that value maximize the function  .
for $p=\frac{6}{25}$ the second derivative is $-125.56<0$ , so $ p=\frac{6}{25} $ gives maximum.

i also got for $p=0,1$ the second derivative is $\quad-\infty$ , so $ p=0,1 $  also give maximum.
  But the book commented out that $p=0,1$ give a minimum.

$\bullet$"How  $p=0,1$ give a minimum?"

There is a typo in the post, presumably you mean $p^6(1-p)^{19}$.
Since $\log$ is an increasing function, finding the maximum of $p^6(1-p)^{19}$ and finding the maximum of its logarithm are equivalent problems. There is a bit of a question mark about the endpoints $p=0$ and $p=1$, but the maximum is clearly not there, so we are examining the logarithm in the interval $(0,1)$, where there is no issue. 
The logarithm is $6\log p +19\log(1-p)$, which has derivative $\frac{6}{p}-\frac{19}{1-p}$. 
This derivative is not $0$ at $p=0$ or $p=1$, in fact it is undefined at these places. 
It is true that the derivative of $p^6(1-p)^{19}$ is $0$ at $p=0$ and $p=1$, in addition to $p=\frac{6}{25}$,  for the derivative is $(p^6)(-19)(1-p)^{18}+(1-p)^{19}(6)(p^5)$. This clearly vanishes at $p=0$ and $p=1$. In fact, by taking out the common factor $p^5(1-p)^{18}$, we can quickly find where the derivative is $0$ without doing the detour through logarithms. 
The second derivative is not really suitable for testing what happens at the endpoints. I assume these are probabilities, so $p$ is restricted to the interval $[0,1]$. Our original expression is obviously $0$ at the endpoints, and positive if $0\lt p\lt 1$, so it is clear that we have an absolute  minimum at $0$ and at $1$.
Actually, I would not use the second derivative test at all. Our derivative simplifies to $\frac{6-25p}{p(1-p)}$. The denominator is positive for all $p$ in the interval $(0,1)$. Looking at the numerator, we can see it is positive up to $p=\frac{6}{25}$ and then negative. So the logarithm is increasing in the interval $(0,\frac{6}{25}]$, and then decreasing, so reaches a maximum at $p=\frac{6}{25}$. 
