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

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(A) $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^{\frac{3}{2}}$બંને બાજુ $\log$ લેતા,આપણને મળે છે$\log y = \frac{3}{2} [\log(x+1) + \log(2x+1) + \log(3x+1)

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

Vedclass · Answer

(A) $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^{\frac{3}{2}}$બંને બાજુ $\log$ લેતા,આપણને મળે છે$\log y = \frac{3}{2} [\log(x+1) + \log(2x+1) + \log(3x+1) + \ldots + \log(nx+1)]$$x$ ની સાપેક્ષમાં વિકલન કરતા,આપણને મળે છે$\frac{1}{y} \frac{dy}{dx} = \frac{3}{2} \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} ight]$$	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]$હવે $x=0$ આગળ,$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}$

જો $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^{\frac{3}{2}}$ હોય,તો $x=0$ આગળ $\frac{dy}{dx}$ ની કિંમત શોધો.

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