What is the derivative of-5e^{-2.5x^2}x ?

The derivative of-5e^{-2.5x^2}x is -5(-5e^{-2.5x^2}x^2+e^{-2.5x^2})

What is the first derivative of-5e^{-2.5x^2}x ?

The first derivative of-5e^{-2.5x^2}x is -5(-5e^{-2.5x^2}x^2+e^{-2.5x^2})

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\frac{d}{d x} (- 5 e^{- 2.5 x^{2}} x)

- 5 (- 5 e^{- 2.5 x^{2}} x^{2} + e^{- 2.5 x^{2}})

Solution steps

\frac{d}{d x} (- 5 e^{- 2.5 x^{2}} x)

Take the constant out:

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

= - 5 \frac{d}{d x} (e^{- 2.5 x^{2}} x)

Apply the Product Rule:

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

f = e^{- 2.5 x^{2}}, g = x

= - 5 (\frac{d}{d x} (e^{- 2.5 x^{2}}) x + \frac{d x}{d x} e^{- 2.5 x^{2}})

\frac{d}{d x} (e^{- 2.5 x^{2}}) = - 5 e^{- 2.5 x^{2}} x

\frac{d x}{d x} = 1

= - 5 ((- 5 e^{- 2.5 x^{2}} x) x + 1 \cdot e^{- 2.5 x^{2}})

Simplify

- 5 ((- 5 e^{- 2.5 x^{2}} x) x + 1 \cdot e^{- 2.5 x^{2}}) : - 5 (- 5 e^{- 2.5 x^{2}} x^{2} + e^{- 2.5 x^{2}})

= - 5 (- 5 e^{- 2.5 x^{2}} x^{2} + e^{- 2.5 x^{2}})

y^{^{'}} = \frac{x - 1}{y - 3}

\int \frac{x ln ( 1 + x ^{2} )}{1 + x ^{2}} d x

\int_{0}^{l n (2)} x e^{x} d x

\frac{\partial}{\partial x} (z \cdot cos (y \cdot x))

y^{^{'}} = 2 y + e^{x}

What is the derivative of-5e^{-2.5x^2}x ?
The derivative of-5e^{-2.5x^2}x is -5(-5e^{-2.5x^2}x^2+e^{-2.5x^2})
What is the first derivative of-5e^{-2.5x^2}x ?
The first derivative of-5e^{-2.5x^2}x is -5(-5e^{-2.5x^2}x^2+e^{-2.5x^2})