If f(x) = sqrt{ax} + frac{a^2}{sqrt{ax}},then | vedclass

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(C) Given $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}} = \sqrt{a} \cdot x^{1/2} + a^2 \cdot a^{-1/2} \cdot x^{-1/2} = \sqrt{a} \cdot x^{1/2} + a^{3/2} \c

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(C) Given $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}} = \sqrt{a} \cdot x^{1/2} + a^2 \cdot a^{-1/2} \cdot x^{-1/2} = \sqrt{a} \cdot x^{1/2} + a^{3/2} \cdot x^{-1/2}$.Differentiating with respect to $x$:$f'(x) = \sqrt{a} \cdot \frac{1}{2} x^{-1/2} + a^{3/2} \cdot \left( -\frac{1}{2} x^{-3/2} \right)$.$f'(x) = \frac{\sqrt{a}}{2\sqrt{x}} - \frac{a^{3/2}}{2x^{3/2}}$.Now,evaluating at $x = a$:$f'(a) = \frac{\sqrt{a}}{2\sqrt{a}} - \frac{a^{3/2}}{2a^{3/2}}$.$f'(a) = \frac{1}{2} - \frac{1}{2} = 0$.

If $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}}$,then $f'(a) = $

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