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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

if we start with \(\displaystyle ln(1+9x)\) and ignore \(\displaystyle x^2\) we can calculate that

\(\displaystyle f'(x)= \frac{9}{1+9x} <=> f'(0)=9\)

\(\displaystyle f''(x)= \frac{9^2}{(1+9x)^2} <=> f'(0)=9^2\)

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\(\displaystyle f^{n}(x)= \frac{9^n}{(1+9x)^n} <=> f'(0)=9^n \)

how does it work after? Don't we have to use product rule?

Regards,

\(\displaystyle |\pi\rangle\)