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

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(B) Given $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$.Substitute $y = \frac{x}{2}$ into the function:$f(x, \frac{x}{2}) = \frac{\cos(x - 4(\frac{x}{

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(B) Given $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$.Substitute $y = \frac{x}{2}$ into the function:$f(x, \frac{x}{2}) = \frac{\cos(x - 4(\frac{x}{2}))}{\cos(x + 4(\frac{x}{2}))} = \frac{\cos(x - 2x)}{\cos(x + 2x)} = \frac{\cos(-x)}{\cos(3x)} = \frac{\cos(x)}{\cos(3x)}$.Now,calculate the partial derivative with respect to $x$:$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( \frac{\cos x}{\cos 3x} \right)$.Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$:$\frac{\partial f}{\partial x} = \frac{(-\sin x)(\cos 3x) - (\cos x)(-3 \sin 3x)}{\cos^2 3x} = \frac{3 \cos x \sin 3x - \sin x \cos 3x}{\cos^2 3x}$.At $y = \frac{x}{2}$,the expression is a function of $x$ only,and its derivative with respect to $x$ is evaluated as shown above.

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

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