Given $$f(x)=\frac{\sqrt{2-e^{2x}}\sqrt[4]{2-e^{4x}}\cdot\ldots\cdot\sqrt[50]{2-e^{50x}}}{(2-e^x)\sqrt[3]{2-e^{3x}}\cdot\ldots\cdot\sqrt[99]{2-e^{99x}}},$$ find $f'(0)$.
this method was used
$$ \Big(\ln|f|\Big)' = \frac{f'}{f}\quad\Rightarrow\quad f'=f\cdot \Big(\ln|f|\Big)' $$
At the end I came across a harmonic series in the power of the number 2: from 1/2 to 1/50 in the numerator and from 1 to 1/99 in the denominator[. no ideas how to convert further
You shouldn’t separate the numerator and denominator when taking $\log$.
We have
$$\ln f =\left(\frac {\ln{(2-e^{2x}})}2 +\frac {\ln{(2-e^{4x}})}4\dots+\frac{\ln(2-e^{50x})}{50}\right)-\left(\ln(2-e^x)+\frac{\ln{(2-e^{3x}})}3+\dots+\frac{\ln(2-e^{99x})}{99}\right)$$
Differentiating, we have $$\frac{f’}f = g(x)=-\left(\frac{e^{2x}}{2-e^{2x}}+\frac{e^{4x}}{2-e^{4x}}+\dots+\frac{e^{50x}}{2-e^{50x}}\right) + \left(\frac {e^x}{2-e^x} +\frac {e^{3x}}{2-e^{3x}}+\dots+\frac{e^{99x}}{2-e^{99x}}\right)$$
And since $f’(0)=f(0)g(0)$ we have $$f’(0)=f(0)\left(50\cdot\frac 1{2-1} - 25\cdot\frac 1{2-1}\right)=25f(0)$$
$$\implies f’(0)=25\left({(2-1)^\frac12(2-1)^\frac14\dots(2-1)^\frac1{50}\over(2-1)(2-1)^\frac13\dots(2-1)^\frac1{99}}\right)=25$$
$$\boxed{f’(0)=25}$$
