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

Question

Dr. Abubacker · Accepted Answer

(b)
$\because f(x,y) = \frac{\cos{(x - 4y)}}{\cos{(x + 4y)}}$
$	herefore f\left( x,\frac{\pi}{2} ight) = \ \frac{\cos{(x - 2\pi)}}{\cos{(x + 2\pi)}} = \frac{\cos x}{\cos x} = 1$
$	herefore\frac{\partial f}{\partial x} = 0$

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

Explanation

Reviewed by Dr. Abubacker

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