What is (\partial)/(\partial z)(e^{-x^2-2y^2-3z^2}) ?

The answer to (\partial)/(\partial z)(e^{-x^2-2y^2-3z^2}) is -6e^{-x^2-2y^2-3z^2}z

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial z)(e^{-x^2-2y^2-3z^2})

\frac{\partial}{\partial z} (e^{- x^{2} - 2 y^{2} - 3 z^{2}})

- 6 e^{- x^{2} - 2 y^{2} - 3 z^{2}} z

Solution steps

\frac{\partial}{\partial z} (e^{- x^{2} - 2 y^{2} - 3 z^{2}})

Treat

x, y

as constants

Apply the chain rule:

e^{- x^{2} - 2 y^{2} - 3 z^{2}} \frac{\partial}{\partial z} (- x^{2} - 2 y^{2} - 3 z^{2})

= e^{- x^{2} - 2 y^{2} - 3 z^{2}} \frac{\partial}{\partial z} (- x^{2} - 2 y^{2} - 3 z^{2})

\frac{\partial}{\partial z} (- x^{2} - 2 y^{2} - 3 z^{2}) = - 6 z

= e^{- x^{2} - 2 y^{2} - 3 z^{2}} (- 6 z)

Simplify

= - 6 e^{- x^{2} - 2 y^{2} - 3 z^{2}} z

\int (3 x + 1)^{11} d x

\int_{0}^{6} x e^{- x} d x

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

(1 + x^{2}) (1 + y^{2}) d x - x y d y = 0

What is (\partial)/(\partial z)(e^{-x^2-2y^2-3z^2}) ?
The answer to (\partial)/(\partial z)(e^{-x^2-2y^2-3z^2}) is -6e^{-x^2-2y^2-3z^2}z