What is (\partial)/(\partial x)(-2e^{-x^2+y}yx) ?

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

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(\partial)/(\partial x)(-2e^{-x^2+y}yx)

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

x, \frac{1}{3} x, 16

y^{^{'}} + \frac{y}{x} = 0

\int \frac{ln ( x + 1 )}{x + 1} d x

\int_{0}^{1} \frac{x ^{3} + 1}{x + 1} d x

ln (\frac{1}{1 + e ^{- x}})

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