Question

lf $a^y={\log }_a(x^2+x+1)$, then $\frac{dy}{dx}=$

• A. $\frac{{\log }_{a}e.(2x+1)}{(x^2+x+1)\log (x^2+x+1)}$
• B. $\frac{(2x+1)}{(x^2+x+1)\log (x^2+x+1)}$
• C. $\frac{1}{(x^2+x+1)\log (x^2+x+1)}$
• D. $\frac{-1}{(x^2+x+1)\log (x^2+x+1)}$

Answer 1

The correct option is A $\frac{{\log }_{a}e.(2x+1)}{(x^2+x+1)\log (x^2+x+1)}$

We have, $a^y={\log }_a(x^2+x+1)\Rightarrow \left(1\right)$
Now differentiating both sides, w.r.t $x$
$\Rightarrow a^y\left(\log a\right)\frac{dy}{dx}={\log }_{a}e\cdot \frac{2x+1}{x^2+x+1}$
$\Rightarrow \frac{dy}{dx}=\frac{{\log }_{a}e\cdot (2x+1)}{(x^2+x+1)\cdot \log (x^2+x+1)}$, using (1)