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

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

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^2})

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

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

Solution steps

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

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}})

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

x \to 1 lim ((x - 2 [x])^{2})

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{- x} - 5 x

f (x) = 3 x^{2} - 10 x

π x^{3}

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

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