Question

if $y=5^{2\left\{{\log }_5\right(x+1)-{\log }_5(3x+1\left)\right\}}$ then $\frac{dy}{dx}$ at $x=0$ is

• A. $0$
• B. $\frac{1}{3}$
• C. $\frac{3}{5}$
• D. $-4$

Answer 1

The correct option is D $-4$

$y=5^{2\left[lo{g}_5\right(x+1)-lo{g}_5(3x+1\left)\right]}$

$y={25}^{\left[o{g}_5\frac{(x+1)}{3x+1}\right]}$

$y={\left(\frac{x+1}{3x+1}\right)}^2$

$=\frac{x^2+1+2x}{9x^2+6x+1}$

${\frac{dy}{dx}|}_{x=0}\frac{(2x+2)(9x^2+6x+1)-(18x+6)(x^2+2x+1)}{(9x^2+6x+1{)}^2}$

$=\frac{(0+2)1-6\left(1\right)}{1}=-4$.