What is the derivative of-16e^{-x^4}x^3 ?

The derivative of-16e^{-x^4}x^3 is -16(-4e^{-x^4}x^6+3e^{-x^4}x^2)

What is the first derivative of-16e^{-x^4}x^3 ?

The first derivative of-16e^{-x^4}x^3 is -16(-4e^{-x^4}x^6+3e^{-x^4}x^2)

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\frac{d}{d x} (- 16 e^{- x^{4}} x^{3})

- 16 (- 4 e^{- x^{4}} x^{6} + 3 e^{- x^{4}} x^{2})

Solution steps

\frac{d}{d x} (- 16 e^{- x^{4}} x^{3})

Take the constant out:

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

= - 16 \frac{d}{d x} (e^{- x^{4}} x^{3})

Apply the Product Rule:

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

f = e^{- x^{4}}, g = x^{3}

= - 16 (\frac{d}{d x} (e^{- x^{4}}) x^{3} + \frac{d}{d x} (x^{3}) e^{- x^{4}})

\frac{d}{d x} (e^{- x^{4}}) = - 4 e^{- x^{4}} x^{3}

\frac{d}{d x} (x^{3}) = 3 x^{2}

= - 16 ((- 4 e^{- x^{4}} x^{3}) x^{3} + 3 x^{2} e^{- x^{4}})

Simplify

- 16 ((- 4 e^{- x^{4}} x^{3}) x^{3} + 3 x^{2} e^{- x^{4}}) : - 16 (- 4 e^{- x^{4}} x^{6} + 3 e^{- x^{4}} x^{2})

= - 16 (- 4 e^{- x^{4}} x^{6} + 3 e^{- x^{4}} x^{2})

\int_{0}^{1} \int_{0}^{y^{2}} x e^{\frac{x}{y ^{2}}} d x d y

x \to 4 lim (\frac{x ^{2} - 3 x - 4}{x - 4})

\frac{d}{d x} (\frac{3 x + 5}{x ^{2} + 4})

x \to 8 lim (4 x)

n = 1 \sum \infty 6 (\frac{5}{4})^{n}

What is the derivative of-16e^{-x^4}x^3 ?
The derivative of-16e^{-x^4}x^3 is -16(-4e^{-x^4}x^6+3e^{-x^4}x^2)
What is the first derivative of-16e^{-x^4}x^3 ?
The first derivative of-16e^{-x^4}x^3 is -16(-4e^{-x^4}x^6+3e^{-x^4}x^2)