What is (\partial)/(\partial y)(y^3e^x-4y+1) ?

The answer to (\partial)/(\partial y)(y^3e^x-4y+1) is 3e^xy^2-4

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(\partial)/(\partial y)(y^3e^x-4y+1)

\frac{\partial}{\partial y} (y^{3} e^{x} - 4 y + 1)

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

Solution steps

\frac{\partial}{\partial y} (y^{3} e^{x} - 4 y + 1)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (y^{3} e^{x}) - \frac{\partial}{\partial y} (4 y) + \frac{\partial}{\partial y} (1)

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

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

\frac{\partial}{\partial y} (1) = 0

= 3 e^{x} y^{2} - 4 + 0

Simplify

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

\int sec^{2} (x) (tan (x) + 3) d x

sin (arcsin (x) + arccos (x))

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

\int (e^{8 x} + e^{- 9 x}) d x

y = ln (3 x^{2} + 2 x - 1)

What is (\partial)/(\partial y)(y^3e^x-4y+1) ?
The answer to (\partial)/(\partial y)(y^3e^x-4y+1) is 3e^xy^2-4