If y = xe^xy then prove that dy/dX = y(1 + xy)/ x(1 - xy)

1.	If y = xe^xy then prove that dy/dX = y(1 + xy)/ x(1 - xy)
Answer» SOLUTION GIVEN $\sf{y = x \: {e}^{xy} \: }$ TO *PROVE* $\displaystyle \sf{ \frac{dy}{dx} = \frac{y(1 + xy)}{x(1 - xy)} \: }$ *PROOF* Here it is given that $\sf{y = x \: {e}^{xy} \: } \: \: ...(1)$ $\displaystyle \sf{ {e}^{xy} = \frac{y}{x} } \: \: \: ....(2)$ Differentiating both sides of Equation (1) with RESPECT to x we get $\displaystyle \sf{ \frac{dy}{dx} = \: {e}^{xy} + x \: {e}^{xy} \bigg( y + x \frac{dy}{dx} \bigg) \: }$ $\implies \displaystyle \sf{ \frac{dy}{dx} = \frac{y}{x} + x \times \frac{y}{x} \times \bigg( y + x \frac{dy}{dx} \bigg) \: }$ ( by EQN. 2) $\implies \displaystyle \sf{ x\frac{dy}{dx} = y + x y \bigg( y + x \frac{dy}{dx} \bigg) \: }$ $\implies \displaystyle \sf{ x\frac{dy}{dx} = y + x {y}^{2} + {x}^{2}y \frac{dy}{dx} \: }$ $\implies \displaystyle \sf{ x\frac{dy}{dx} - {x}^{2}y \frac{dy}{dx} = y + x {y}^{2} \: }$ $\implies \displaystyle \sf{ \big(x - {x}^{2} y \big)\frac{dy}{dx}= y + x {y}^{2} \: }$ $\implies \displaystyle \sf{ \frac{dy}{dx}= \frac{y + x {y}^{2} }{x - {x}^{2} y} \: }$ $\implies \displaystyle \sf{ \frac{dy}{dx}= \frac{y(1 + x {y}^{} )}{x(1 - {x}^{} y)} \: }$ *Hence* *proved* *━━━━━━━━━━━━━━━━* *LEARN MORE FROM BRAINLY* The VALUE of x for the minimum The value of x for the minimumvalue of √3 cos x +sin x is brainly.in/question/24003780

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