What is the derivative of y=(ln(x))^{ln(x)} ?

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

What is the first derivative of y=(ln(x))^{ln(x)} ?

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

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derivative of

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

x^{l n (l n (x)) - 1} (ln (ln (x)) + 1)

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (\frac{z}{x + y})

\frac{d y}{d x} = \frac{( 2 x + 5 )}{( x ^{2} + 2 x )}, y (1) = 1

\int 2 x^{\frac{3}{2}} d x

\frac{1}{x}, \frac{1}{x ^{2}}, 1, 2

\frac{d}{d x} (\frac{x ^{3} - x}{1 + x ^{2}})

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