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

The answer to (\partial)/(\partial x)(2xy^2-2x^3y) is 2y^2-6yx^2

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

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

2 y^{2} - 6 y x^{2}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (2 x^{3} y) = 6 y x^{2}

= 2 y^{2} - 6 y x^{2}

n = 0 \sum \infty \frac{( ( 1 ) ^{n} )}{n}

d er i v a t i v e y = \frac{x ^{2} + 1}{2 x ^{2} - 1}

\frac{d}{d x} (1 - cos (x - 1))

t an g e n t f (x) = 12 x - 2 x, at x = 9

\int \frac{3}{9 + x ^{2}} d x

What is (\partial)/(\partial x)(2xy^2-2x^3y) ?
The answer to (\partial)/(\partial x)(2xy^2-2x^3y) is 2y^2-6yx^2