What is (\partial)/(\partial y)(2xy^4e^y+2xy^3+y) ?

The answer to (\partial)/(\partial y)(2xy^4e^y+2xy^3+y) is 2x(4y^3e^y+e^yy^4)+6xy^2+1

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(\partial)/(\partial y)(2xy^4e^y+2xy^3+y)

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

2 x (4 y^{3} e^{y} + e^{y} y^{4}) + 6 x y^{2} + 1

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (2 x y^{3}) = 6 x y^{2}

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

= 2 x (4 y^{3} e^{y} + e^{y} y^{4}) + 6 x y^{2} + 1

f (x) = 19 - x^{2}, at x = 3

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

\frac{d}{d x} (\frac{- y}{x})

(- 3, 7), (2, - 5)

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

What is (\partial)/(\partial y)(2xy^4e^y+2xy^3+y) ?
The answer to (\partial)/(\partial y)(2xy^4e^y+2xy^3+y) is 2x(4y^3e^y+e^yy^4)+6xy^2+1