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

The answer to (\partial)/(\partial y)(e^{-x^2-2y^2-3z^2}) is -4e^{-x^2-2y^2-3z^2}y

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

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

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

Solution steps

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

Treat

x, z

as constants

Apply the chain rule:

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

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

\frac{\partial}{\partial y} (- x^{2} - 2 y^{2} - 3 z^{2}) = - 4 y

= e^{- x^{2} - 2 y^{2} - 3 z^{2}} (- 4 y)

Simplify

= - 4 e^{- x^{2} - 2 y^{2} - 3 z^{2}} y

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What is (\partial)/(\partial y)(e^{-x^2-2y^2-3z^2}) ?
The answer to (\partial)/(\partial y)(e^{-x^2-2y^2-3z^2}) is -4e^{-x^2-2y^2-3z^2}y