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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

\int 8 sec (x) (sec (x) - 4 tan (x)) d x

y^{^{''}} + 8 y^{^{'}} + 4 y = t

\frac{\partial}{\partial y} (sin (z))

4 x + 9, 144 - x, - 9, 144

\frac{2}{x ^{3}} - \frac{8}{x}

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