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

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(B) दिया गया है $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$.हमें $\left. \frac{\partial f}{\partial x} \right|_{y = \frac{x}{4}}$ का मान ज्ञात करना

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(B) दिया गया है $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$.हमें $\left. \frac{\partial f}{\partial x} ight|_{y = \frac{x}{4}}$ का मान ज्ञात करना है.सबसे पहले,$f$ का $x$ के सापेक्ष आंशिक अवकलन करने पर:$\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)}$.अब $y = \frac{x}{4}$ रखने पर:$\left. \frac{\partial f}{\partial x} ight|_{y = \frac{x}{4}} = \frac{\sin(8(\frac{x}{4}))}{\cos^2(x + 4(\frac{x}{4}))} = \frac{\sin(2x)}{\cos^2(2x)} = 	an(2x) \sec(2x)$.

यदि $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$ है,तो $\left. \frac{\partial f}{\partial x} \right|_{y = \frac{x}{4}}$ का मान ज्ञात कीजिए:

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