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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{( y + 1 )}{y} d y

f (x) = \frac{e ^{x}}{x ^{2} + 6}

\int \frac{3 x - 1}{( x + 4 ) ^{2}} d x

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

\int x^{3} e^{a x^{4}} d x

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