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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\frac{\partial}{\partial y} (6 x y^{2} - 6 x^{2} y + 5 x y)

x^{4} - 5 x^{2} + 4

\frac{d}{d x} (e^{s i n (2 x)} + sin (e^{2 x}))

x \to 2 lim (\frac{1}{x ( x - 2 )})