What is the first (ln^2(x))^' ?

The first (ln^2(x))^' is (2ln(x))/x

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(ln^2(x))^'

(ln^{2} (x))^{^{'}}

\frac{2 ln ( x )}{x}

Solution steps

(ln^{2} (x))^{'}

Apply the chain rule:

2 ln (x) (ln (x))^{'}

= 2 ln (x) (ln (x))^{'}

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

= 2 ln (x) \frac{1}{x}

Simplify

2 ln (x) \frac{1}{x} : \frac{2 ln ( x )}{x}

= \frac{2 ln ( x )}{x}

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

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

\int 2 x (x^{2} + 2)^{6} d x

\frac{\partial}{\partial x} (3 x^{4} y^{2} - 2)

n = 1 \sum \infty \frac{1}{( n ) ( n + 8 )}