(x + 1)2 + (x + 2)3 + (x + 3)4


(+ 1)2 + (+ 2)3 + (+ 3)4


Let $\quad y=(x+1)^{2}(x+2)^{3}(x+3)^{4}$

Taking log on both sides,

$\log y=\log \left[(x+1)^{2} \cdot(x+2)^{3} \cdot(x+3)^{4}\right]$

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

$\lceil\because \log x y=\log x+\log$

$\Rightarrow \log y=2 \log (x+1)+3 \log (x+2)+4 \log (x+3)$

$\left[\because \log x^{y}=y \log x\right]$

Differentiating both sides w.r.t. $x$;

$\frac{1}{y} \cdot \frac{d y}{d x}=2 \cdot \frac{d}{d x} \log (x+1)+3 \cdot \frac{d}{d x} \log (x+2)+4 \cdot \frac{d}{d x} \log (x+3)$

$\frac{1}{y} \cdot \frac{d y}{d x}=2 \cdot \frac{1}{x+1}+3 \cdot \frac{1}{x+2}+4 \cdot \frac{1}{x+3}$

$\frac{d y}{d x}=y\left[\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3}\right]$


$\Rightarrow \quad \frac{d y}{d x}=(x+1)^{2}(x+2)^{3}(x+3)^{4}\left[\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3}\right]$



$=(x+1)(x+2)^{2}(x+3)^{3}\left(2 x^{2}+10 x+12+3 x^{2}+12 x+9\right.$ $\left.+4 x^{2}+12 x+8\right)$

$=(x+1)(x+2)^{2}(x+3)^{3}\left(9 x^{2}+34 x+29\right)$

Thus, $\frac{d y}{d x}=(x+1)(x+2)^{2}(x+3)^{3}\left(9 x^{2}+34 x+29\right)$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now