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Question Number 96231 by joki last updated on 30/May/20

Commented by prakash jain last updated on 30/May/20

$$x^{3}y-3{xy}^3=2a^4$$$$\frac{d}{dx}(x^{3}y-3{xy}^3)=\frac{d}{dx}2a^4$$$$3x^{2}y+x^3\frac{dy}{dx}-3y^3-9{xy}^2\frac{dy}{dx}=0$$$$\frac{dy}{dx}(x^3-9{xy}^2)=3y^3-3x^{2}y$$$$\frac{dy}{dx}=\frac{3x^{2}y-3y^3}{9{xh}^2-x^3}$$

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