What is (\partial)/(\partial x)(-6e^{-3x^2-y^2}x) ?

The answer to (\partial)/(\partial x)(-6e^{-3x^2-y^2}x) is -6(-6e^{-3x^2-y^2}x^2+e^{-3x^2-y^2})

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(\partial)/(\partial x)(-6e^{-3x^2-y^2}x)

\frac{\partial}{\partial x} (- 6 e^{- 3 x^{2} - y^{2}} x)

- 6 (- 6 e^{- 3 x^{2} - y^{2}} x^{2} + e^{- 3 x^{2} - y^{2}})

Solution steps

\frac{\partial}{\partial x} (- 6 e^{- 3 x^{2} - y^{2}} x)

Treat

y

as a constant

Take the constant out:

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

= - 6 \frac{\partial}{\partial x} (e^{- 3 x^{2} - y^{2}} x)

Apply the Product Rule:

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

f = e^{- 3 x^{2} - y^{2}}, g = x

= - 6 (\frac{\partial}{\partial x} (e^{- 3 x^{2} - y^{2}}) x + \frac{\partial}{\partial x} (x) e^{- 3 x^{2} - y^{2}})

\frac{\partial}{\partial x} (e^{- 3 x^{2} - y^{2}}) = - 6 e^{- 3 x^{2} - y^{2}} x

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

= - 6 ((- 6 e^{- 3 x^{2} - y^{2}} x) x + 1 \cdot e^{- 3 x^{2} - y^{2}})

Simplify

- 6 ((- 6 e^{- 3 x^{2} - y^{2}} x) x + 1 \cdot e^{- 3 x^{2} - y^{2}}) : - 6 (- 6 e^{- 3 x^{2} - y^{2}} x^{2} + e^{- 3 x^{2} - y^{2}})

= - 6 (- 6 e^{- 3 x^{2} - y^{2}} x^{2} + e^{- 3 x^{2} - y^{2}})

(1 - 2 x) (3 x - 4), at x = 2

f (x) = 2 x^{2} - x + 9

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

\frac{d}{d x} (ln (ln (sec (2 x))))

y = 3 x + \frac{4}{x}

What is (\partial)/(\partial x)(-6e^{-3x^2-y^2}x) ?
The answer to (\partial)/(\partial x)(-6e^{-3x^2-y^2}x) is -6(-6e^{-3x^2-y^2}x^2+e^{-3x^2-y^2})