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If $f(x)=\sqrt{a x}+\frac{a^2}{\sqrt{a x}}$, then $f^{\prime}(a)$ is equal to
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Verified Answer
The correct answer is:
$0$
We have,
$$
f(x)=\sqrt{a x}+\frac{a^2}{\sqrt{a x}}
$$
On differentiating w.r.t. $x$, we get,
$$
\begin{aligned}
& f^{\prime}(x)=\sqrt{a} \cdot \frac{1}{2 \sqrt{x}}+\frac{a^2}{\sqrt{a}} \cdot\left(-\frac{1}{2 x \sqrt{x}}\right) \\
& f^{\prime}(x)=\frac{\sqrt{a}}{2} \cdot \frac{1}{\sqrt{x}}-\frac{a \sqrt{a}}{2 x \sqrt{x}} \\
& f^{\prime}(a)=\frac{\sqrt{a}}{2} \cdot \frac{1}{\sqrt{a}}-\frac{a \sqrt{a}}{2 a \sqrt{a}}=\frac{1}{2}-\frac{1}{2}=0
\end{aligned}
$$
$$
f(x)=\sqrt{a x}+\frac{a^2}{\sqrt{a x}}
$$
On differentiating w.r.t. $x$, we get,
$$
\begin{aligned}
& f^{\prime}(x)=\sqrt{a} \cdot \frac{1}{2 \sqrt{x}}+\frac{a^2}{\sqrt{a}} \cdot\left(-\frac{1}{2 x \sqrt{x}}\right) \\
& f^{\prime}(x)=\frac{\sqrt{a}}{2} \cdot \frac{1}{\sqrt{x}}-\frac{a \sqrt{a}}{2 x \sqrt{x}} \\
& f^{\prime}(a)=\frac{\sqrt{a}}{2} \cdot \frac{1}{\sqrt{a}}-\frac{a \sqrt{a}}{2 a \sqrt{a}}=\frac{1}{2}-\frac{1}{2}=0
\end{aligned}
$$
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