## University Calculus: Early Transcendentals (3rd Edition)

$-\frac{2}{|x^2+1|\sqrt( {( {x^2})+2})}$
Given that $y$ = $cosec^{-1}(x^2+1)$ where $x\gt0$ on substituting ${x^2+1}$ = $u$ on applying chain rule derivative $\frac{dy}{dx}$ = $\frac{{dcosec^{-1}{u}}}{du}$ $\times$ $\frac{du}{dx}$ or $\frac{dy}{dx}$ = $-\frac{1}{{|u|}\sqrt ({u^2}-1)}$ $\times$ $\frac{du}{dx}$ so $\frac{dy}{dx}$ = $-\frac{1}{|(x^2+1)|\sqrt ({({x^2+1})^2-1})}$ $\times$ $\frac{d {(x^2+1)}}{dx}$ or $\frac{dy}{dx}$ = $-\frac{1}{|x^2+1|\sqrt( {({x^4})+2x^2})}$ $\times$ $2x$ The final answer is: $-\frac{2}{|x^2+1|\sqrt( {( {x^2})+2})}$