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

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

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

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

3 y^{2}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (3 y^{2}) = 0

= 3 y^{2} - 0

Simplify

= 3 y^{2}

\int (x^{2} - 3 x) e^{2 x} d x

x \to 0 - lim (\frac{6 x}{x ^{4}})

y^{^{''}} - 4 y^{^{'}} - 5 y = 4 e^{- 2 t}, y (0) = 0, y^{^{'}} (0) = - 1

\int sin^{4} (3 t) cos^{4} (3 t) d t

\int_{0}^{1} - 1 x 5^{x} d x

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