What is the derivative of-1/2 e^{-x/4} ?

The derivative of-1/2 e^{-x/4} is 1/8 e^{-x/4}

What is the first derivative of-1/2 e^{-x/4} ?

The first derivative of-1/2 e^{-x/4} is 1/8 e^{-x/4}

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\frac{d}{d x} (- \frac{1}{2} e^{- \frac{x}{4}})

\frac{1}{8} e^{- \frac{x}{4}}

Solution steps

\frac{d}{d x} (- \frac{1}{2} e^{- \frac{x}{4}})

Take the constant out:

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

= - \frac{1}{2} \frac{d}{d x} (e^{- \frac{x}{4}})

Apply the chain rule:

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

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

\frac{d}{d x} (- \frac{x}{4}) = - \frac{1}{4}

= - \frac{1}{2} e^{- \frac{x}{4}} (- \frac{1}{4})

Simplify

- \frac{1}{2} e^{- \frac{x}{4}} (- \frac{1}{4}) : \frac{1}{8} e^{- \frac{x}{4}}

= \frac{1}{8} e^{- \frac{x}{4}}

x \to 0 + lim (x lo g_{e} (x))

- 8 (64 e^{- 4 x^{2}} x^{3} - 24 e^{- 4 x^{2}} x)

\frac{t ^{2}}{2}

\int e^{- s t} (2 t + 1) d t

y = \frac{1}{x ^{2}}, y = x, y = \frac{1}{8} x

What is the derivative of-1/2 e^{-x/4} ?
The derivative of-1/2 e^{-x/4} is 1/8 e^{-x/4}
What is the first derivative of-1/2 e^{-x/4} ?
The first derivative of-1/2 e^{-x/4} is 1/8 e^{-x/4}