If $y=\left|\tan x-\left|\sin x\right|\right|$, then the value of $\frac{dy}{dx}$ at $x=\frac{5\pi }{4}$ is

In the neighbourhood of $x=\frac{5\pi }{4}, \sin x<0$
$\therefore \left|\sin x\right|=-\sin x$
Now, $y=\left|\tan x-(-\sin x)\right|=\left|\tan x+\sin x\right|$
Now, in the neighbourhood of $x=\frac{5\pi }{4}$
$\tan x+\sin x\simeq 1+\left(-\frac{1}{\sqrt{2}}\right)$, which is positive
$\therefore$ $y=\tan x+\sin x$
Hence, $\frac{dy}{dx}={\sec }^{2}x+\cos x$
Putting $x=\frac{5\pi }{4}$, we get,

$\frac{dy}{dx}={\left(-\sqrt{2}\right)}^2-\frac{1}{\sqrt{2}}=\frac{2\sqrt{2}-1}{\sqrt{2}}$