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

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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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$\frac{3}{5}$

$\frac{1}{3}$

$- 4$

answer is D.

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$y = 5^{2 \{\log_{5} (x + 1) - \log_{5} (3 x + 1)\}}$

$= 5^{\log_{5} {(\frac{x + 1}{3 x + 1})}^{2}} = {(\frac{x + 1}{3 x + 1})}^{2}$

$\frac{d y}{d x} = 2 {(\frac{x + 1}{3 x + 1})}^{} \times \frac{- 2}{{(3 x + 1)}^{2}}$

$\Rightarrow a t x = 0, \frac{d y}{d x} = - 4$

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