If y = [(x+1)(2x+1)(3x+1) ldots (nx+1)]^{frac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^{\frac{3}{2}}$Taking $\log$ on both sides,we get$\log y = \frac{3}{2} [\log(x+1) + \log(2x+1) + \log(3x+1)

Vedclass · Accepted Answer

$\frac{3n(n+1)}{4}$

Vedclass · Answer

(A) $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^{\frac{3}{2}}$Taking $\log$ on both sides,we get$\log y = \frac{3}{2} [\log(x+1) + \log(2x+1) + \log(3x+1) + \ldots + \log(nx+1)]$Differentiating with respect to $x$,we get$\frac{1}{y} \frac{dy}{dx} = \frac{3}{2} \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} ight]$$	herefore \frac{dy}{dx} = \frac{3y}{2} \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} ight]$Now at $x=0$,$y = [(1)(1)(1) \ldots (1)]^{\frac{3}{2}} = 1$$\left. \frac{dy}{dx} ight|_{x=0} = \frac{3(1)}{2} \left[ \frac{1}{0+1} + \frac{2}{0+1} + \frac{3}{0+1} + \ldots + \frac{n}{0+1} ight]$$= \frac{3}{2} (1 + 2 + 3 + \ldots + n)$$= \frac{3}{2} 	imes \frac{n(n+1)}{2}$$= \frac{3n(n+1)}{4}$

If $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^{\frac{3}{2}}$,then $\frac{dy}{dx}$ at $x=0$ is

Similar Questions

The derivative of $f(x)=x^{\tan ^{-1} x}$ with respect to $g(x)=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ is

If $f(x)=\frac{\cos ^2 x}{1+\sin ^2 x}$, then $f\left(\frac{\pi}{4}\right)-3 f^{\prime}\left(\frac{\pi}{4}\right)=$

If $y = x^x$,then $\frac{dy}{dx} = $

$\text{If } \frac{d}{dx} \left( \frac{x^2+1}{(x^2+5)(x^2+9)} \right) = \frac{2x(x^2+1)}{(x^2+5)(x^2+9)} \left[ \frac{1}{f(x)} - \frac{1}{g(x)} - \frac{1}{h(x)} \right], \text{ then } 2h(x) - f(x) - g(x) = $

If $y = \sin^{-1}\left[\cos \sqrt{\frac{1+x}{2}}\right] + x^x$,then find $\frac{dy}{dx}$ at $x = 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module