Finding the derivative of $y = x^{(x+1)(x+2)(x+3)(x+4)\ldots(x+n)}$.

Question

I'm trying to find the derivative of this function with respect to $x$:

$$y = x^{(x+1)(x+2)(x+3)(x+4)\ldots(x+n)}$$

I was thinking about using $\ln$ to solve this, but I'm not sure if that's the right way to go about it. Any ideas or suggestions would be greatly appreciated.

Thanks for your help!

Answer 1

\begin{align}y&=x^{(x+1)(x+2)\cdots(x+n)}\\\ln y&=(x+1)(x+2)\cdots(x+n)\ln x\\\ln(\ln y)&=\ln(x+1)+\ln(x+2)+\cdots+\ln(x+n)+\ln(\ln x)\\\dfrac1{\ln y}\cdot\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}&=\dfrac1{x+1}+\dfrac1{x+2}+\cdots\dfrac1{x+n}+\dfrac1{\ln x}\cdot\dfrac1x\\\dfrac{\mathrm dy}{\mathrm dx}&=y\ln y\left(\sum_{i=1}^n\dfrac1{x+i}+\dfrac1{x\ln x}\right)\\&=x^{(x+1)(x+2)\cdots(x+n)}(x+1)(x+2)\cdots(x+n)\ln x\left(\sum_{i=1}^n\dfrac1{x+i}+\dfrac1{x\ln x}\right)\end{align}

Answer 2

$\displaystyle y\ =\ x^{(( x+1)( x+2) ...( x+n)}$
Taking log on both sides: $\displaystyle \log( y) \ =\ ( x+1)( x+2) ...( x+n) \log( x)$
Differentiating this: \begin{align} \displaystyle \frac{dy}{dx} &= y\left(\left(\sum _{k=1}^{n}\frac{( x+1) ( x+2) ( x+3) \cdots( x+n)}{( x+k)}\right) \log x \\\qquad +\ \frac{( x+1) ( x+2) ( x+3) \cdots( x+n)}{x}\right) \end{align}