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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

x \to - \infty lim (b x)

\frac{d}{d x} (ln (sec (x) + tan (x)))

\int 1200 e^{0.3 t} d t

\int_{0}^{1} \frac{1}{9 + 16 x ^{2}} d x

\frac{x}{( 3 - x ) ^{3}}

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