Question
Find \(y'(0)\) if \(y = (x+1) (x+2) (x+3) … (x+2023)\)
Solution:
\(\ln y = \ln (x+1) + \ln (x+2) + \ln (x+3) + ….. + \ln (x+2023)\)
Differentiating with respect to \(x\)
\(\frac{d(\ln y)}{dx} = \frac{\ln (x+1) + \ln (x+2) + \ln (x+3) + ….. + \ln (x+2023)}{dx}\)
\(\Rightarrow \frac{1}{y}\cdot\frac{dy}{dx} = \frac{1}{x+1}\cdot(0 + 1) + \frac{1}{x+2}\cdot(0 + 1) + \frac{1}{x+3}\cdot(0 + 1) + …. + \frac{1}{x+2023}\cdot(0 + 1)\)
\(\Rightarrow \frac{dy}{dx} = y'(x) = y(\frac{1}{x+1} + \frac{1}{x+2} + \frac{1}{x+3} + ….. + \frac{1}{x+2023})\)
\(\Rightarrow y'(x) = (x+1)(x+2)(x+3)…..(x+2023)(\frac{1}{x+1} + \frac{1}{x+2} + \frac{1}{x+3} + ….. + \frac{1}{x+2023})\)
Therefore \(y'(0) = 1\cdot 2 \cdot 3 ….. 2023 (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ….. + \frac{1}{2023})\)
\(\Rightarrow y'(0) = 2023! \cdot (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ….. + \frac{1}{2023})\)