What is (\partial)/(\partial y)(2xe^{x^2-y^2}) ?

The answer to (\partial)/(\partial y)(2xe^{x^2-y^2}) is -4e^{x^2-y^2}xy

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(\partial)/(\partial y)(2xe^{x^2-y^2})

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

\int y e^{- y} d y

\frac{d}{d x} (cos^{2} (sin (4 x)))

y^{^{''}} + 5 y^{^{'}} + 6 y = 2 e^{x}, y (0) = 1, y^{^{'}} (0) = - 1

\int 4 sin (5 x) + 5 cos (3 x) d x

\frac{d y}{d t} = 2 y - t

What is (\partial)/(\partial y)(2xe^{x^2-y^2}) ?
The answer to (\partial)/(\partial y)(2xe^{x^2-y^2}) is -4e^{x^2-y^2}xy