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Vlad1618 [11]
3 years ago
9

How do you calculate the derivative of y with respect to x on the equation x^3y+3xy^3=x+y?

Mathematics
1 answer:
lesya692 [45]3 years ago
3 0
\bf x^3y+3xy^3=x+y\\\\
-----------------------------\\\\
\left[ 3x^2y+x^3\frac{dy}{dx} \right]+3\left[ 1\cdot y^3+x\cdot 3y^2\frac{dy}{dx} \right]=1+\frac{dy}{dx}
\\\\\\
3x^2y+x^3\frac{dy}{dx}+3y^3+9xy^2\frac{dy}{dx}-\frac{dy}{dx}=1\impliedby \textit{common factor}
\\\\\\
\cfrac{dy}{dx}(x^3+9xy^2-1)+3x^2y+3y^3=1
\\\\\\
\cfrac{dy}{dx}(x^3+9xy^2-1)=1-3x^2y-3y^3
\\\\\\
\cfrac{dy}{dx}=\cfrac{1-3x^2y-3y^3}{x^3+9xy^2-1}
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