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

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

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

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

3 x y^{2} + 4 e^{x} - 4 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 3 x y^{2} + 4 e^{x} - 4 y

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

f (x) = - 3 x^{3} - 2 x^{2} - x + 3, at x = - 1

\int \frac{x}{( 1 - x ^{2} ) ^{2}} d x

\int_{1}^{2} sin (x) + tan (x) d x

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