Question

Find $\frac{dy}{dx}$
$y=\frac{1}{x^2-2x+3}$

Answer 1

We have,

$y=\frac{1}{x^2-2x+3}$

On differentiating w.r.t $x$, we get

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

$\frac{dy}{dx}=\frac{(x^2-2x+3)\times 0-(2x-2+0)}{(x^2-2x+3{)}^2}$

$\frac{dy}{dx}=\frac{0-(2x-2)}{(x^2-2x+3{)}^2}$

$\frac{dy}{dx}=\frac{(2-2x)}{(x^2-2x+3{)}^2}$

Hence, this is the answer.