What is the derivative of f(x)=-5e^{-2.5x^2}x ?

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

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

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

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derivative of

f (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}})

x \to 0 lim (x^{3} + 2^{2} - x)

x y \frac{d y}{d x} = 1 + y^{2}

y^{^{'}} - \frac{2}{x} y = \frac{y ^{5}}{x ^{2}}

\frac{d}{d x} (\frac{lo g _{e} ( x )}{x ^{2}})

y^{^{''}} - 2 y^{^{'}} - 3 y = 0

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