What is (\partial)/(\partial x)(2xy^3-4x^2y^2) ?

The answer to (\partial)/(\partial x)(2xy^3-4x^2y^2) is 2y^3-8y^2x

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(\partial)/(\partial x)(2xy^3-4x^2y^2)

\frac{\partial}{\partial x} (2 x y^{3} - 4 x^{2} y^{2})

2 y^{3} - 8 y^{2} x

Solution steps

\frac{\partial}{\partial x} (2 x y^{3} - 4 x^{2} 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^{3}) - \frac{\partial}{\partial x} (4 x^{2} y^{2})

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

\frac{\partial}{\partial x} (4 x^{2} y^{2}) = 8 y^{2} x

= 2 y^{3} - 8 y^{2} x

x^{\frac{1}{2}} + y^{\frac{1}{2}} = 4

f (x) = arcsin (4 x)

f (x) = - 6 x^{2} - 10 x, at x = 1

\frac{t ^{2} + 1}{t ^{2} - 1}

y = x (2 x + 3)^{5}

What is (\partial)/(\partial x)(2xy^3-4x^2y^2) ?
The answer to (\partial)/(\partial x)(2xy^3-4x^2y^2) is 2y^3-8y^2x