What is (\partial)/(\partial x)(4xye^{x-y^2}) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(4xye^{x-y^2})

\frac{\partial}{\partial x} (4 x y e^{x - y^{2}})

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

Solution steps

\frac{\partial}{\partial x} (4 x y e^{x - y^{2}})

Treat

y

as a constant

Take the constant out:

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

= 4 y \frac{\partial}{\partial x} (x e^{x - y^{2}})

Apply the Product Rule:

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

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

= 4 y (\frac{\partial}{\partial x} (x) e^{x - y^{2}} + \frac{\partial}{\partial x} (e^{x - y^{2}}) x)

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

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

= 4 y (1 \cdot e^{x - y^{2}} + e^{x - y^{2}} x)

Simplify

= 4 y (e^{x - y^{2}} + e^{x - y^{2}} x)

x \to 0 lim (e^{x^{2}} - 1)

\frac{d y}{d x} = x^{2} - y - 2, y (0) = 1

\int 5 x^{\frac{3}{2}} + 2 x d x

\int (1 - 2 x^{2} - \frac{4}{x ^{- 3}}) x^{- 3} d x

\int (7 x^{2} - 12 x) e^{2 x} d x

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