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

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(A) Given the function: $f(x) = \sqrt{ax} + a^2(ax)^{-1/2}$.Differentiating with respect to $x$ using the chain rule:$f^{\prime}(x) = \frac{d}{dx}(\sq

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(A) Given the function: $f(x) = \sqrt{ax} + a^2(ax)^{-1/2}$.Differentiating with respect to $x$ using the chain rule:$f^{\prime}(x) = \frac{d}{dx}(\sqrt{a} \cdot x^{1/2}) + \frac{d}{dx}(a^2 \cdot \sqrt{a}^{-1} \cdot x^{-1/2})$$f^{\prime}(x) = \sqrt{a} \cdot \frac{1}{2} x^{-1/2} + a^2 \cdot \frac{1}{\sqrt{a}} \cdot (-\frac{1}{2}) x^{-3/2}$$f^{\prime}(x) = \frac{\sqrt{a}}{2\sqrt{x}} - \frac{a^2}{2\sqrt{a} \cdot x\sqrt{x}}$Substituting $x = a$ into the derivative:$f^{\prime}(a) = \frac{\sqrt{a}}{2\sqrt{a}} - \frac{a^2}{2\sqrt{a} \cdot a\sqrt{a}}$$f^{\prime}(a) = \frac{1}{2} - \frac{a^2}{2a^2} = \frac{1}{2} - \frac{1}{2} = 0$.

If $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}}$,then $f^{\prime}(a)$ is equal to

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