What is the derivative of-x^{-x} ?

The derivative of-x^{-x} is -x^{-x}(-ln(x)-1)

What is the first derivative of-x^{-x} ?

The first derivative of-x^{-x} is -x^{-x}(-ln(x)-1)

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

- x^{- x} (- ln (x) - 1)

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{- x} = e^{(- x) l n (x)}

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

Apply the chain rule:

e^{(- x) l n (x)} \frac{d}{d x} ((- x) ln (x))

= e^{(- x) l n (x)} \frac{d}{d x} ((- x) ln (x))

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

= - e^{(- x) l n (x)} (- ln (x) - 1)

Simplify

- e^{(- x) l n (x)} (- ln (x) - 1) : - x^{- x} (- ln (x) - 1)

= - x^{- x} (- ln (x) - 1)

\frac{\partial}{\partial y} (cos (x) cos (y) + 2 x)

\frac{d}{d x} (2 y + x e^{y})

\frac{d y}{d t} - \frac{2}{t} y = t^{5}

\int (\frac{1}{x} - \frac{5}{x ^{2} + 1}) d x

\frac{s + 2}{s ( s + 2 ) + 1}

What is the derivative of-x^{-x} ?
The derivative of-x^{-x} is -x^{-x}(-ln(x)-1)
What is the first derivative of-x^{-x} ?
The first derivative of-x^{-x} is -x^{-x}(-ln(x)-1)