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If $y=\ell \mathrm{n} \sqrt{\tan x}$, then what is the value of $\frac{d y}{d x}$ at $x=\frac{\pi}{4}$?
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Let $\mathrm{y}=\ln \sqrt{\tan x}$
Differentiate both side w.r.t $^{\prime} x^{\prime}$
$\frac{d y}{d x}=\frac{1}{\sqrt{\tan x}} \cdot \frac{1}{2 \sqrt{\tan x}} \cdot \sec ^{2} x$
Now, $\frac{d y}{d x}$ at $x=\pi / 4$
$=\frac{1}{\sqrt{\tan \frac{\pi}{4}}} \times \frac{1}{2 \sqrt{\tan \frac{\pi}{4}}} \times \frac{1}{\cos ^{2}(\pi / 4)}$
$=\frac{1}{2} \times 1 \times \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^{2}}=\frac{1}{2} \times 1 \times 2=1$
Differentiate both side w.r.t $^{\prime} x^{\prime}$
$\frac{d y}{d x}=\frac{1}{\sqrt{\tan x}} \cdot \frac{1}{2 \sqrt{\tan x}} \cdot \sec ^{2} x$
Now, $\frac{d y}{d x}$ at $x=\pi / 4$
$=\frac{1}{\sqrt{\tan \frac{\pi}{4}}} \times \frac{1}{2 \sqrt{\tan \frac{\pi}{4}}} \times \frac{1}{\cos ^{2}(\pi / 4)}$
$=\frac{1}{2} \times 1 \times \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^{2}}=\frac{1}{2} \times 1 \times 2=1$
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