derivative of many roots if $$ y=\frac{(1+2x)^{1/2}.(1+4x)^{1/4}.(1+6x)^{1/6} ... (1+100x)^{1/100}}{(1+3x)^{1/3}.(1+5x)^{1/5}.(1+7x)^{1/7} ... (1+101x)^{1/101}}$$
then find y' at x=0


*

*Already tried to find a general term that I can differentiate.

*Tried using the Pi notation. But I'm not very familiar with it.

 A: Consider $\ln y=...$
Then $\frac 1y\frac{dy}{dx}=...?$
A: Hint: Compute $\dfrac{y'(x)}{y(x)}$ and use the fact that for
$f = f_1 f_2 ... f_n$ 
$$
 \frac{f'}{f} = \frac{f_1'}{f_1} + \frac{f_2'}{f_2} + ... + \frac{f_n'}{f_n}
$$
(which follows from the product rule for the derivative).
A: When you face such monsters, you can be sure of two things


*

*No human being would do it using the product and quotient rules for derivatives

*There must be a trick and, as already said in comments and answers, logarithmic differentiation is more then useful when you face products or quotients.


Using the product notation, the numerator can write $$A=\prod_{k=1}^n (1+2 k x)^{\frac{1}{2 k}}$$ and the denominator $$B=\prod_{k=1}^n (1+(2 k +1)x)^{\frac{1}{2 k+1}}$$ So,  $$y=\frac AB\implies\log(y)=\log(A)-\log(B)=\sum_{k=1}^n \frac{\log(1+2kx)}{2k}-\sum_{k=1}^n \frac{\log(1+(2k+1)x)}{2k+1}$$ Now, compute the derivative $$\frac {y'}y=\sum_{k=1}^n \frac 1{1+2kx}-\sum_{k=1}^n \frac 1{1+(2k+1)x}$$ Now, you want the value of the derivative for $x=0$; this gives the beautiful $\frac {y'}y=0$ and then $y'=0$ whatever $n$ could be.
A: Note: This answer is a variation based upon quotient and product rule.

Let $y(x)=\frac{f(x)}{g(x)}$ with 
  \begin{align*}
f(x)=\prod_{k=1}^{50}u_{2k}(x)\qquad\qquad g(x)=\prod_{k=1}^{50}u_{2k+1}(x)
\end{align*}
  and 
  \begin{align*}
u_k(x)=(1+kx)^{\frac{1}{k}}\qquad 1\leq k\leq 101
\end{align*}
We observe $u_k(0)=1$ for all $k$ and conclude
  \begin{align*}
f(0)=g(0)=1\tag{1}
\end{align*}

We want to calculate $y^\prime(x)$ at $x=0$. So, we consider the quotient rule
\begin{align*}
y^{\prime}(x)=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^2(x)}\tag{2}
\end{align*}

The evaluation at $x=0$ gives according to (1) 
  \begin{align*}
y^{\prime}(0)=\frac{f^{\prime}(0)\cdot 1- 1\cdot g^{\prime}(0)}{1}=f^{\prime}(0)- g^{\prime}(0)\tag{3}
\end{align*}
The derivative of $u_k$ is
  \begin{align*}
u_k^{\prime}(x)=(1+kx)^{\frac{1}{k}-1}\qquad 1\leq k \leq 101
\end{align*}
  and we obtain $u_k^{\prime}(0)=1$.

To calculate $y^{\prime}(0)$ we need $f^{\prime}(x)$ and $g^{\prime}(x)$ evaluated at zero.

We obtain according to the general Leibniz Product Rule
  \begin{align*}
f^{\prime}(x)&=\sum_{j=1}^{50}{u^{\prime}_{2j}(x)}\prod_{{k=1}\atop{k\neq j}}^{50}u_{2k}(x)\\
g^{\prime}(x)&=\sum_{j=1}^{50}{u^{\prime}_{2j+1}(x)}\prod_{{k=1}\atop{k\neq j}}^{50}u_{2k+1}(x)\\
\end{align*}
  and since $u_k(0)=u^{\prime}_k(0)=1$ we conclude $f^{\prime}(0)=g^{\prime}(0)=1$.
We finally obtain according to (3)
  \begin{align*}
y^{\prime}(0)=1-1=0
\end{align*}

