If $f(x) = \sqrt {(ax)} + \dfrac{{{a^2}}}{{\sqrt {ax} }}$ , then $f'(a) = $
A. $ - 1$
B. $0$
C. $1$
D. $a$

Hint:Since the problem is based on the differentiation of a function and we know that,$f'(x)$ represents the first derivative of a function with respect to $x$ hence, we will first calculate $f'(x)$ of a given function in the problem and then calculate the answer by substituting the value of $x = a$ in the last step of the solution.

Formula Used: The formula used in this problem is the derivative of ${x^n}$ which is given as:
$\dfrac{d}{{dx}}({x^n}) = n \cdot {x^{n - 1}}$

Complete step by step Solution:
$f(x) = \sqrt {(ax)} + \dfrac{{{a^2}}}{{\sqrt {ax} }}$ (given)
Let $y = f(x)$ and we can also write the given function as:
$y = f(x) = \sqrt {a} \mathrm{x}^{\frac{1}{2}}+\dfrac{a^2}{\sqrt{a}}\mathrm{x}^{\frac{-1}{2}}$ … (1)
We know that, $\dfrac{d}{{dx}}({x^n}) = n.{x^{n - 1}}$
Now, differentiate the equation (1) with respect to $x$, we get
\[\dfrac{{dy}}{{dx}} = f'(x) = \sqrt a .\left( {\dfrac{1}{2}} \right){x^{\left( {\dfrac{1}{2} - 1} \right)}} + \dfrac{{{a^2}}}{{\sqrt a }}.\left( { - \dfrac{1}{2}} \right){x^{\left( { - \dfrac{1}{2} - 1} \right)}}\]
\[\dfrac{{dy}}{{dx}} = f'(x) = \left( {\dfrac{{\sqrt a }}{2}} \right){x^{\left( { - \dfrac{1}{2}} \right)}} - \left( {\dfrac{{a\sqrt a }}{2}} \right){x^{\left( { - \dfrac{3}{2}} \right)}}\]
Taking common \[\dfrac{{\sqrt a }}{2}\] from the above expression we get
i.e., \[f'(x) = \dfrac{{\sqrt a }}{2}\left( {\dfrac{1}{{\sqrt x }} - \dfrac{a}{{x\sqrt x }}} \right) = \dfrac{{\sqrt a }}{{2\sqrt x }}\left( {1 - \dfrac{a}{x}} \right)\] … (2)
Now, substitute the value $x = a$ in equation (2) we get
\[f'(a) = \dfrac{{\sqrt a }}{{2\sqrt a }}\left( {1 - \dfrac{a}{a}} \right)\]
Thus, the value of \[f'(a) = 0\]

Hence, the correct option is B.

Note: In the problems based on differentiation, apply the required properties(formula) to get an accurate solution to the problem. Sometimes, the problem seems to be complex in differentiation hence, it is advised to break the problem into little parts and then calculations must be performed.