Question

y²cotx= x²coty. find out dy/ dx​

Answer 1

${y}^{2} cotx = {x}^{2} coty \\ differentiate \: w.r.t \: \: \: \: x \\ cotx \: 2y \frac{dy}{dx} + {y}^{2} ( - {cosec}^{2} x) = {x}^{2} ( - {cosec}^{2} y) \frac{dy}{dx} + coty2x \\ \frac{dy}{dx} 2ycotx + {x}^{2} {cosec}^{2} y \frac{dy}{dx} = {y}^{2} {cosec}^{2} x + 2xcoty \\ \frac{dy}{dx} = \frac{ {y}^{2} {cosec}^{2} x + 2xcoty }{2ycotx + {x}^{2} {cosec}^{2} y }$