Question - In Pic

My progress is that i found out that $\displaystyle f''(x) = \frac{[f'(x)]^2}{f(x)}$

but i couldn't get to the solution

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- Feb 28th 2011, 06:03 AMcupidProof of double derivative.
**Question - In Pic**

My progress is that i found out that $\displaystyle f''(x) = \frac{[f'(x)]^2}{f(x)}$

but i couldn't get to the solution - Feb 28th 2011, 06:21 AMNOX Andrew
If you use the identity:

$\displaystyle \cos{x} + i\sin{x} = e^{ix}$,

then

$\displaystyle y = e^{ix} \cdot e^{3ix} \cdot ... \cdot e^{\left(2n-1\right)ix}$

$\displaystyle = e^{ix + 3ix + ... + \left(2n-1\right)ix}$

$\displaystyle = e^{ix\left(1 + 3 + ... + 2n-1\right)$

$\displaystyle = e^{n^2ix}$ - Feb 28th 2011, 06:34 AMcupid
Thanks a lot Andrew

that made the question soo simple

thanks a lot