Consider the function $$f(x)=\frac{(1+2x)^\frac{1}{2}(1+4x)^\frac{1}{4}\cdot\cdot\cdot(1+2nx)^\frac{1}{2n}}{(1+3x)^\frac{1}{3}(1+5x)^\frac{1}{5}\cdot\cdot\cdot(1+(2m+1)x)^\frac{1}{2m+1}}$$ where $n$ and $m$ are positive integers satisfying $n+m-20=f'(0)=2010$. Then $n=$?
Not sure how to approach this question.
$f(x)$ is given in the form $$f(x)=\dfrac{\displaystyle\prod_{j=1}^n (1+2jx)^{1/2j}}{\displaystyle\prod_{k=1}^m\Big(1+(2k+1)x\Big)^{1/(2k+1)}}\\ \implies \ln (f(x))=\sum_{j=1}^n \dfrac1{2j}\ln\Big(1+2jx\Big)-\sum_{k=1}^m \dfrac{1}{2k+1}\ln\Big(1+(2k+1)x\Big)$$ and differentiating with respect to $x$ on both sides, you should get $$\dfrac{f'(x)}{f(x)}=\sum_{j=1}^n \dfrac1{2j}\dfrac{2j}{(1+2jx)}-\sum_{k=1}^m\dfrac{1}{2k+1}\dfrac{2k+1}{(1+(2k+1)x)} \qquad (1)$$ You already have $n+m=2010+20$, what do you get when you put $x=0$ in $(1)$? Value of $f'(0)$ is given to you, and $f(0)=?\cdots$
The final result should be $n=2020$.