What is (\partial)/(\partial x)(2xe^{x^2+y^2}) ?

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

8 y^{^{''}} + 32 y^{^{'}} = 0, y (2) = 1, y^{^{'}} (- 1) = 1

\frac{x ^{2}}{2 + x}

\frac{d x}{d y} = (4 x + 2 y + 1)

y = (8 x^{2} - 16 x + 16) e^{x}

\frac{d}{d x} (\frac{4 x}{x + 1})

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