What is (\partial)/(\partial x)(xye^{-8(x^2+y^2)}) ?

The answer to (\partial)/(\partial x)(xye^{-8(x^2+y^2)}) is y(e^{-8(x^2+y^2)}-16e^{-8(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)(xye^{-8(x^2+y^2)})

\frac{\partial}{\partial x} (x y e^{- 8 (x^{2} + y^{2})})

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

Solution steps

\frac{\partial}{\partial x} (x y e^{- 8 (x^{2} + y^{2})})

Treat

y

as a constant

Take the constant out:

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

= y \frac{\partial}{\partial x} (x e^{- 8 (x^{2} + y^{2})})

Apply the Product Rule:

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

f = x, g = e^{- 8 (x^{2} + y^{2})}

= y (\frac{\partial}{\partial x} (x) e^{- 8 (x^{2} + y^{2})} + \frac{\partial}{\partial x} (e^{- 8 (x^{2} + y^{2})}) x)

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

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

= y (1 \cdot e^{- 8 (x^{2} + y^{2})} + (- 16 e^{- 8 (x^{2} + y^{2})} x) x)

Simplify

y (1 \cdot e^{- 8 (x^{2} + y^{2})} + (- 16 e^{- 8 (x^{2} + y^{2})} x) x) : y (e^{- 8 (x^{2} + y^{2})} - 16 e^{- 8 (x^{2} + y^{2})} x^{2})

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

\int_{0}^{1} 8 cos (2 x) d x

\int_{0}^{9} \frac{1}{9 - x} d x

\frac{d}{d x} (4 x^{\frac{5}{4}} + 10 x^{\frac{3}{2}} + 10 x)

\frac{d}{d x} ((x - 4) \cdot a x^{2} + ln (\frac{x}{4}) \cdot b x^{4})

\int cos (8 t) d t

What is (\partial)/(\partial x)(xye^{-8(x^2+y^2)}) ?
The answer to (\partial)/(\partial x)(xye^{-8(x^2+y^2)}) is y(e^{-8(x^2+y^2)}-16e^{-8(x^2+y^2)}x^2)