Question

diffrentiation (x^2+y^2)^2=xy

Answer 1

$(x^2+y^2)^2=xy\\ \\we\ know\ \frac{d}{dx}f(g(x,y))=f'(g(x,y))*g'(x,y)\\ \\2(x^2+y^2)^{2-1} . \frac{d}{dx}(x^2+y^2) = x . \frac{dy}{dx} + y . 1\\ \\2(x^2+y^2) . (2 x + 2 y \frac{dy}{dx})=x. \frac{dy}{dx}+y\\ \\4(x^2+y^2)x+4(x^2+y^2)y.\frac{dy}{dx}-x.\frac{dy}{dx}=y\\ \\4(x^2+y^2)y.\frac{dy}{dx}-x.\frac{dy}{dx}=y-4(x^2+y^2)x\\ \\ (4x^2y+4y^3-x).\frac{dy}{dx}=y-4(x^3+y^2x)\\ \\\frac{dy}{dx}=\frac{y-4x^3-4y^2x}{(4x^2y+4y^3-x)}$

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$Let\ w = y^2\\.\ \ \ \frac{dw}{dx}=\frac{dw}{dy} * \frac{dy}{dx} =2y^{2-1}*\frac{dy}{dx}\\ \\Let\ z=(x^2+y^2)\\ \\\frac{dz}{dx} = \frac{d(x^2)}{dx}+\frac{d(y^2)}{dx}=2x+\frac{d(y^2)}{dx}=2x+\frac{d(y^2}{dy}*\frac{dy}{dx}=2x+2y\frac{dy}{dx}\\ \\let\ f=z^2\\ \\\frac{df}{dx}=\frac{d(z^2)}{dz}*\frac{dz}{dx}=2 z \frac{dz}{dx}= 2(x^2+y^2)(2x+2y\frac{dy}{dx})$