What is the derivative of 1+ln^2(x) ?

The derivative of 1+ln^2(x) is (2ln(x))/x

What is the first derivative of 1+ln^2(x) ?

The first derivative of 1+ln^2(x) is (2ln(x))/x

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

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

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (1) + \frac{d}{d x} (ln^{2} (x))

\frac{d}{d x} (1) = 0

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

= 0 + \frac{2 ln ( x )}{x}

Simplify

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

\frac{\partial}{\partial x} (tan (2 x))

\int_{0}^{1} 2 π x (x - x^{3}) d x

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

\frac{d}{d x} (\frac{2 x ^{9} - 5 x + x}{x ^{4}})

\int_{0}^{r} 0 d r

What is the derivative of 1+ln^2(x) ?
The derivative of 1+ln^2(x) is (2ln(x))/x
What is the first derivative of 1+ln^2(x) ?
The first derivative of 1+ln^2(x) is (2ln(x))/x