What is the derivative of f(x)=[ln(x)]^x ?

The derivative of f(x)=[ln(x)]^x is ln^x(x)(ln(ln(x))+1/(ln(x)))

What is the first derivative of f(x)=[ln(x)]^x ?

The first derivative of f(x)=[ln(x)]^x is ln^x(x)(ln(ln(x))+1/(ln(x)))

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derivative of

f (x) = [ln (x)]^{x}

ln^{x} (x) (ln (ln (x)) + \frac{1}{ln ( x )})

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

= e^{x l n (l n (x))} (ln (ln (x)) + \frac{1}{ln ( x )})

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

= ln^{x} (x) (ln (ln (x)) + \frac{1}{ln ( x )})

y^{'} = 0.0015 y (2100 - y)

\frac{d y}{d x} = x^{\frac{3}{2}}

\int - \frac{1}{n ^{2}} cos (n t) d t

\frac{\partial}{\partial x} (x^{3} cos (x y))

x = y^{2}, x = 6 - 2 y^{2}

What is the derivative of f(x)=[ln(x)]^x ?
The derivative of f(x)=[ln(x)]^x is ln^x(x)(ln(ln(x))+1/(ln(x)))
What is the first derivative of f(x)=[ln(x)]^x ?
The first derivative of f(x)=[ln(x)]^x is ln^x(x)(ln(ln(x))+1/(ln(x)))