wonder if anyone know what the derivate to y = ln(x + sqrt(1+x^2)) is ?? :/
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y´ = 1/sqrt(x^2+1) right ?
no it is not rite $\displaystyle \displaystyle ln\left( x + \sqrt{1+x^2}\right) $ in general if $\displaystyle y=\displaystyle ln\left(f(x) \right) $ $\displaystyle y' = \frac{f'(x)}{f(x)} $ so...
(1/sqrt(x^2+1)) / (ln(x + sqrt(1+x^2))) ?
Originally Posted by paulaa (1/sqrt(x^2+1)) / (ln(x + sqrt(1+x^2))) ? no again ... $\displaystyle \dfrac{d}{dx} \left[\ln(x+\sqrt{1+x^2})\right] = \dfrac{1+\frac{x}{\sqrt{1+x^2}}}{x + \sqrt{1+x^2}}$
So she was right the first time! (Which figures, for the log form of arcsinh.)
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