Question

If $y=\frac{1}{1+x+x^2}$, then $\frac{dy}{dx}$ is equal to

• A. $y^2(1+2x)$
• B. $\frac{-\left(1+2x\right)}{y^2}$
• C. $\frac{(1+2x)}{2}$
• D. $-y(1+2x)$
• E. $-y^2(1+2x)$

Answer 1

Answer: (E)

Given, $y=\frac{1}{1+x+x^2}$
$\Rightarrow 1+x+x^2=\frac{1}{y}$
On differentiating w.r.t. $x$, we get
$0+1+2x=-\frac{1}{y^2}\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=-(1+2x)y^2$