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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

f (x) = e^{x} \cdot cos (x), at x = 0

\int z^{m} d z

\frac{1}{1 - x}, 0

\int \frac{x}{1 - x} d x

y^{^{'}} + y = x

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