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

The answer to (\partial)/(\partial x)(2x^2-2xy^2) is 4x-2y^2

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

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

4 x - 2 y^{2}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (2 x^{2}) = 4 x

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

= 4 x - 2 y^{2}

\int \frac{x + 3}{x - 1} d x

\frac{d}{d x} (arcsin (\frac{x}{π}))

\int_{0}^{4} x^{2} + 1 d x

\frac{1 + s \cdot 2}{( 1 + s \cdot 3 ) ^{2}}

f (x) = x^{2} sin (\frac{1}{x})

What is (\partial)/(\partial x)(2x^2-2xy^2) ?
The answer to (\partial)/(\partial x)(2x^2-2xy^2) is 4x-2y^2