If $y=[(x+1)(2 x+1)(3 x+1) \ldots \ldots(\mathrm{n} x+1)]^{\frac{3}{2}}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=0$ is

Question

Aicharya · Accepted Answer

$y=[(x+1)(2 x+1)(3 x+1) \ldots(\mathrm{n} x+1)]^{\frac{3}{2}}$ Taking 'log' on both sides, we get $\begin{array}{r} \log y=\frac{3}{2}[\log (x+1)+\log (2 x+1)+\log (3 x+1) \ +\ldots+\log (n x+1)] \end{array}$ Differentiating w.r.t. $x$, we get $\begin{aligned} & \quad \frac{1}{y} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3}{2}\left[\frac{1}{x+1}+\frac{2}{2 x+1}+\frac{3}{3 x+1}+\ldots+\frac{\mathrm{n}}{\mathrm{n} x+1}ight] \ & 	herefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 y}{2}\left[\frac{1}{x+1}+\frac{2}{2 x+1}+\frac{3}{3 x+1}+\ldots+\frac{\mathrm{n}}{\mathrm{n} x+1}ight] \ & 	ext { Now at } x=0, y=[\underbrace{(1)(1)(1) \ldots(1)}_{\mathrm{n} 	ext { times }}]^{\frac{3}{2}}=1 \end{aligned}$ $\begin{aligned}\left.	herefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}ight|_{x=0} & =\frac{3(1)}{2}\left[\frac{1}{0+1}+\frac{2}{0+1}+\frac{3}{0+1}+\ldots+\frac{\mathrm{n}}{0+1}ight] \ & =\frac{3}{2}(1+2+3+\ldots+\mathrm{n}) \ & =\frac{3}{2} 	imes \frac{\mathrm{n}(\mathrm{n}+1)}{2} \ & =\frac{3 \mathrm{n}(\mathrm{n}+1)}{4}\end{aligned}$

If $y=[(x+1)(2 x+1)(3 x+1) \ldots \ldots(\mathrm{n} x+1)]^{\frac{3}{2}}$, then $\frac{\mathrm{d}…

Solution