What is (\partial)/(\partial y)(-e^{-xy^2}*y^2) ?

The answer to (\partial)/(\partial y)(-e^{-xy^2}*y^2) is 2e^{-xy^2}xy^3-2e^{-xy^2}y

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(\partial)/(\partial y)(-e^{-xy^2}*y^2)

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{- x y^{2}}, g = y^{2}

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

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

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

= - ((- 2 x e^{- x y^{2}} y) y^{2} + 2 y e^{- x y^{2}})

Simplify

- ((- 2 x e^{- x y^{2}} y) y^{2} + 2 y e^{- x y^{2}}) : 2 e^{- x y^{2}} x y^{3} - 2 e^{- x y^{2}} y

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

\frac{\partial}{\partial x} (e^{s i n (x^{2})})

\int \frac{5 arcsin ( x )}{x ^{2}} d x

x^{3} y^{^{'}} + 2 x^{2} y = - 3 cos (3 x) y^{- \frac{1}{2}}

y^{^{'}} + 2 x y - 6 = 0

\frac{d}{d x} (\frac{2 x}{4})

What is (\partial)/(\partial y)(-e^{-xy^2}*y^2) ?
The answer to (\partial)/(\partial y)(-e^{-xy^2}*y^2) is 2e^{-xy^2}xy^3-2e^{-xy^2}y