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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{'}} = 2 y - 1

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

\int \frac{49 - x ^{2}}{x ^{4}} d x

\frac{d ^{3}}{d x ^{3}} (x^{6} e^{x})

f (x) = \frac{6 x + 5}{x - 1}, at x = 2

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