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If $f(x, y)=\frac{\cos (x-4 y)}{\cos (x+4 y)}$, then $\left.\frac{\partial f}{\partial x}\right|_{y=\frac{x}{2}}$ is equal to :
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$\because \quad f(x, y)=\frac{\cos (x-4 y)}{\cos (x+4 y)}$
$\therefore \quad f\left(x, \frac{\pi}{2}\right)=\frac{\cos (x-2 \pi)}{\cos (x+2 \pi)}$
$=\frac{\cos (2 \pi-x)}{\cos (2 \pi+x)}=\frac{\cos x}{\cos x}=1$
$\therefore \quad \frac{\partial f}{\partial x}=0$
$\therefore \quad f\left(x, \frac{\pi}{2}\right)=\frac{\cos (x-2 \pi)}{\cos (x+2 \pi)}$
$=\frac{\cos (2 \pi-x)}{\cos (2 \pi+x)}=\frac{\cos x}{\cos x}=1$
$\therefore \quad \frac{\partial f}{\partial x}=0$
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