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

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(A) Given function is $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}}$.We can rewrite the function as $f(x) = \sqrt{a} \cdot x^{1/2} + a^2 \cdot \sqrt{a}^{-

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(A) Given function is $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}}$.We can rewrite the function as $f(x) = \sqrt{a} \cdot x^{1/2} + a^2 \cdot \sqrt{a}^{-1} \cdot x^{-1/2} = \sqrt{a} \cdot x^{1/2} + a^{3/2} \cdot x^{-1/2}$.Differentiating with respect to $x$ using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:$f^{\prime}(x) = \sqrt{a} \cdot \frac{1}{2} x^{-1/2} + a^{3/2} \cdot \left(-\frac{1}{2}\right) x^{-3/2}$.$f^{\prime}(x) = \frac{\sqrt{a}}{2\sqrt{x}} - \frac{a^{3/2}}{2x\sqrt{x}}$.Now, substitute $x = a$ into the derivative:$f^{\prime}(a) = \frac{\sqrt{a}}{2\sqrt{a}} - \frac{a^{3/2}}{2a\sqrt{a}}$.$f^{\prime}(a) = \frac{1}{2} - \frac{a^{3/2}}{2a^{3/2}}$.$f^{\prime}(a) = \frac{1}{2} - \frac{1}{2} = 0$.

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

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