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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

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

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

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

Simplify

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

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

y^{^{''}} + y = 5 e^{- 7 t}

e^{- 5 t} sin (2 t)

\int - x^{2} + 4 x d x

(2 x y - sec^{2} (x)) d x + (x^{2} + 2 y) d y = 0

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