If f(x, y) = frac{cos(x - 4y)}{cos(x + 4y)},t | vedclass

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(B) Given $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$.We need to evaluate $\left. \frac{\partial f}{\partial x} \right|_{y = \frac{x}{4}}$.First,sub

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(B) Given $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$.We need to evaluate $\left. \frac{\partial f}{\partial x} ight|_{y = \frac{x}{4}}$.First,substitute $y = \frac{x}{4}$ into the function:$f\left(x, \frac{x}{4}ight) = \frac{\cos(x - 4(\frac{x}{4}))}{\cos(x + 4(\frac{x}{4}))} = \frac{\cos(x - x)}{\cos(x + x)} = \frac{\cos(0)}{\cos(2x)} = \frac{1}{\cos(2x)} = \sec(2x)$.Now,differentiate $f$ with respect to $x$:$\frac{d}{dx} [\sec(2x)] = 2 \sec(2x) 	an(2x)$.However,if the question implies evaluating the partial derivative first and then substituting $y = \frac{x}{4}$:$\frac{\partial f}{\partial x} = \frac{\cos(x+4y)(-\sin(x-4y)) - \cos(x-4y)(-\sin(x+4y))}{\cos^2(x+4y)}$$= \frac{\sin(x+4y)\cos(x-4y) - \cos(x+4y)\sin(x-4y)}{\cos^2(x+4y)} = \frac{\sin((x+4y) - (x-4y))}{\cos^2(x+4y)} = \frac{\sin(8y)}{\cos^2(x+4y)}$.Substituting $y = \frac{x}{4}$:$\left. \frac{\partial f}{\partial x} ight|_{y = \frac{x}{4}} = \frac{\sin(8(\frac{x}{4}))}{\cos^2(x + 4(\frac{x}{4}))} = \frac{\sin(2x)}{\cos^2(2x)} = 	an(2x) \sec(2x)$.

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

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