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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

\int_{2}^{6} \frac{x}{x ^{2} + 6 x + 13} d x

\int - \frac{1}{x ^{2} + 1} d x

y = x^{7}, - 2, 3

t \to 0 lim (\frac{8}{t} - \frac{8}{t ^{2} + t})

tan (y)

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