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0 1 If it's from to then that's indeed a classic magic differentiation exercise! But here's an alternative:
0 1 If it's from to then that's indeed a classic magic differentiation exercise! But here's an alternative:
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\displaystyle \frac{d}{d\lambda}\left( \frac{x^{\lambda}-1}{\log{x}} \right) = \frac{d}{d\lambda}\left( \frac{x^{\lambda}}{\log{x}} \right) = \frac{1}{\log{x}}\frac{d}{d \lambda}\left x^{ \lambda}
\displaystyle \frac{d}{d \lambda}\left x^{ \lambda} = x^{\lambda}\log{x}
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