What is (\partial)/(\partial x)(2xy-x^3-y^2) ?

The answer to (\partial)/(\partial x)(2xy-x^3-y^2) is 2y-3x^2

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(\partial)/(\partial x)(2xy-x^3-y^2)

\frac{\partial}{\partial x} (2 x y - x^{3} - y^{2})

2 y - 3 x^{2}

Solution steps

\frac{\partial}{\partial x} (2 x y - x^{3} - y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (2 x y) - \frac{\partial}{\partial x} (x^{3}) - \frac{\partial}{\partial x} (y^{2})

\frac{\partial}{\partial x} (2 x y) = 2 y

\frac{\partial}{\partial x} (x^{3}) = 3 x^{2}

\frac{\partial}{\partial x} (y^{2}) = 0

= 2 y - 3 x^{2} - 0

Simplify

= 2 y - 3 x^{2}

\int t (e^{- s t}) d t

\frac{3}{2} t + 2 e^{- t} - \frac{1}{4} e^{- 2 t} - \frac{7}{4}

\int \frac{3 x + 1}{( x - 2 ) ( x + 3 ) ^{2}} d x

\frac{\partial}{\partial x} (\frac{( y + z )}{x})

\frac{d}{d x} (k (sec (x) + tan (x)))

What is (\partial)/(\partial x)(2xy-x^3-y^2) ?
The answer to (\partial)/(\partial x)(2xy-x^3-y^2) is 2y-3x^2