What is (\partial)/(\partial x)(-10xe^{-5x^2-2y^2}) ?

The answer to (\partial)/(\partial x)(-10xe^{-5x^2-2y^2}) is -10(e^{-5x^2-2y^2}-10e^{-5x^2-2y^2}x^2)

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(\partial)/(\partial x)(-10xe^{-5x^2-2y^2})

\frac{\partial}{\partial x} (- 10 x e^{- 5 x^{2} - 2 y^{2}})

- 10 (e^{- 5 x^{2} - 2 y^{2}} - 10 e^{- 5 x^{2} - 2 y^{2}} x^{2})

Solution steps

\frac{\partial}{\partial x} (- 10 x e^{- 5 x^{2} - 2 y^{2}})

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - 10 \frac{\partial}{\partial x} (x e^{- 5 x^{2} - 2 y^{2}})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = e^{- 5 x^{2} - 2 y^{2}}

= - 10 (\frac{\partial}{\partial x} (x) e^{- 5 x^{2} - 2 y^{2}} + \frac{\partial}{\partial x} (e^{- 5 x^{2} - 2 y^{2}}) x)

\frac{\partial}{\partial x} (x) = 1

\frac{\partial}{\partial x} (e^{- 5 x^{2} - 2 y^{2}}) = - 10 e^{- 5 x^{2} - 2 y^{2}} x

= - 10 (1 \cdot e^{- 5 x^{2} - 2 y^{2}} + (- 10 e^{- 5 x^{2} - 2 y^{2}} x) x)

Simplify

- 10 (1 \cdot e^{- 5 x^{2} - 2 y^{2}} + (- 10 e^{- 5 x^{2} - 2 y^{2}} x) x) : - 10 (e^{- 5 x^{2} - 2 y^{2}} - 10 e^{- 5 x^{2} - 2 y^{2}} x^{2})

= - 10 (e^{- 5 x^{2} - 2 y^{2}} - 10 e^{- 5 x^{2} - 2 y^{2}} x^{2})

\int 6 x + 2 d x

- 3 - 3 x^{2}

f (x) = \frac{3 - 2 x - x ^{2}}{x ^{2} - 1}

\frac{d y}{d x} = x (1 + y)

\frac{\partial}{\partial x} (2 x y + y^{2} + 1)

What is (\partial)/(\partial x)(-10xe^{-5x^2-2y^2}) ?
The answer to (\partial)/(\partial x)(-10xe^{-5x^2-2y^2}) is -10(e^{-5x^2-2y^2}-10e^{-5x^2-2y^2}x^2)