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

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Aicharya · Accepted Answer

We have, $ f(x)=\sqrt{a x}+\frac{a^2}{\sqrt{a x}} $ On differentiating w.r.t. $x$, we get, $ \begin{aligned} & f^{\prime}(x)=\sqrt{a} \cdot \frac{1}{2 \sqrt{x}}+\frac{a^2}{\sqrt{a}} \cdot\left(-\frac{1}{2 x \sqrt{x}}ight) \ & f^{\prime}(x)=\frac{\sqrt{a}}{2} \cdot \frac{1}{\sqrt{x}}-\frac{a \sqrt{a}}{2 x \sqrt{x}} \ & f^{\prime}(a)=\frac{\sqrt{a}}{2} \cdot \frac{1}{\sqrt{a}}-\frac{a \sqrt{a}}{2 a \sqrt{a}}=\frac{1}{2}-\frac{1}{2}=0 \end{aligned} $

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

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