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

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

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

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

3 x^{2} + 2 x y^{3}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

Simplify

= 3 x^{2} + 2 x y^{3}

f (x) = (6 x^{2} - 9 x) e^{x}

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

n = 1 \sum \infty (\frac{1}{6})^{n}

x^{2} + 8 x - 7

\frac{9 x}{2 ^{x}}

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