Question

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

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

Answer 1

The correct option is E $-y^2(1+2x)$

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