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

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

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

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

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

Solution steps

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

Simplify

e^{- (x^{2} + y^{2} + 2 x)} : e^{- x^{2} - y^{2} - 2 x}

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

\int_{2}^{4} x^{2} e^{2 x} d x

(e^{e x} - x^{2} y) = 1

\int (\frac{x ^{2} - 1}{x ^{4}}) d x

(x^{2} + 5) y^{^{'}} = x y

y = 3 x^{2}, (0, 1)

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