If y=x tan y,then frac{d y}{d x}= | vedclass

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(B) Given equation: $y = x 	an y$ Differentiating both sides with respect to $x$: $\frac{d y}{d x} = x \sec^2 y \frac{d y}{d x} + 	an y$ Rearranging

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$\frac{y}{x-x^2-y^2}$

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(B) Given equation: $y = x 	an y$ Differentiating both sides with respect to $x$: $\frac{d y}{d x} = x \sec^2 y \frac{d y}{d x} + 	an y$ Rearranging the terms to isolate $\frac{d y}{d x}$: $\frac{d y}{d x} - x \sec^2 y \frac{d y}{d x} = 	an y$ $\frac{d y}{d x} (1 - x \sec^2 y) = 	an y$ $\frac{d y}{d x} = \frac{	an y}{1 - x \sec^2 y}$ Multiply numerator and denominator by $x$: $\frac{d y}{d x} = \frac{x 	an y}{x - x^2 \sec^2 y}$ Since $y = x 	an y$,we have $	an y = \frac{y}{x}$: $\frac{d y}{d x} = \frac{y}{x - x^2 (1 + 	an^2 y)} = \frac{y}{x - x^2 - x^2 	an^2 y}$ Substituting $x^2 	an^2 y = y^2$: $\frac{d y}{d x} = \frac{y}{x - x^2 - y^2}$

If $y=x \tan y$,then $\frac{d y}{d x}=$

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