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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

= - 2 (1 \cdot e^{x^{2} - y^{2}} + (- 2 e^{x^{2} - y^{2}} y) y)

Simplify

- 2 (1 \cdot e^{x^{2} - y^{2}} + (- 2 e^{x^{2} - y^{2}} y) y) : - 2 (e^{x^{2} - y^{2}} - 2 e^{x^{2} - y^{2}} y^{2})

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

\int (1 + \frac{9 x}{8}) d x

x y^{^{'}} + y = - 2 x y^{2}, y (1) = - 2

f (x) = x^{3} + 4 x^{2} + 5 x, at x = - 1

\int \frac{1}{4 x ^{2} + 12 x + 16} d x

(7, - 5), (- 7, - 5)

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