If y=52(log5(x+1)−log5(3x+1)) then dydx at x=0 is

y=52{log5(x+13x+1)} =5log5(x+13x+1)2 y=(x+1)2(3x+1)2 y=x2+2x+19x2+6x+1 (dydx)x=0=(9x2+6x+1)(2x+2)−(x2+2x+1)(18x+6)(9x2+6x+1)2 =1(2)−1(6)1 = - 4

If y=52(log5(x+1)−log5(3x+1)) then dydx at x=0 is

y=52{log5(x+13x+1)} =5log5(x+13x+1)2 y=(x+1)2(3x+1)2 y=x2+2x+19x2+6x+1 (dydx)x=0=(9x2+6x+1)(2x+2)−(x2+2x+1)(18x+6)(9x2+6x+1)2 =1(2)−1(6)1 = - 4

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If $y = 5^{2 (\log_{5} (x + 1) - \log_{5} (3 x + 1))}$ then $\frac{d y}{d x}$ at $x = 0$ is

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$\frac{3}{5}$

-4

$\frac{1}{3}$

answer is D.

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Detailed Solution

$y = 5^{2 {\log_{5} (\frac{x + 1}{3 x + 1})}} = 5^{\log_{5} {(\frac{x + 1}{3 x + 1})}^{2}} y = \frac{{(x + 1)}^{2}}{{(3 x + 1)}^{2}} y = \frac{x^{2} + 2 x + 1}{9 x^{2} + 6 x + 1} {(\frac{d y}{d x})}_{x = 0} = \frac{(9 x^{2} + 6 x + 1) (2 x + 2) - (x^{2} + 2 x + 1) (18 x + 6)}{{(9 x^{2} + 6 x + 1)}^{2}} = \frac{1 (2) - 1 (6)}{1} = - 4$