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

The (ln(x^2+1))^' is (2x)/(x^2+1)

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

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

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(ln (x^{2} + 1))^{^{'}}

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

Solution steps

(ln (x^{2} + 1))^{'}

Apply the chain rule:

\frac{1}{x ^{2} + 1} (x^{2} + 1)^{'}

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

(x^{2} + 1)^{'} = 2 x

= \frac{1}{x ^{2} + 1} \cdot 2 x

Simplify

\frac{1}{x ^{2} + 1} \cdot 2 x : \frac{2 x}{x ^{2} + 1}

= \frac{2 x}{x ^{2} + 1}

\frac{d x}{d t} = k x (k - x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (- 1) = 0, y^{^{'}} (- 1) = 1

\int \frac{3 x + 5}{x ( 2 x - 1 )} d x

\int e^{3 x} sin (5 x) d x

t^{3} e^{t} t