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

The answer to (\partial)/(\partial y)(x-y^3+y^2sin(x)) is -3y^2+2ysin(x)

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(\partial)/(\partial y)(x-y^3+y^2sin(x))

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

- 3 y^{2} + 2 y sin (x)

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (x) = 0

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

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

= 0 - 3 y^{2} + 2 y sin (x)

Simplify

= - 3 y^{2} + 2 y sin (x)

e^{(\frac{2}{x})}

\int_{- 2}^{2} - 3 x^{2} - 4 x d x

\int \frac{5}{x ^{2} + 3 x - 4} d x

\int (5 x - 3)^{4} d x

y = \frac{x ^{2}}{3} + x + \frac{1}{2}

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