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

$\frac {-2s^2}{\sqrt{1-s^2}}$
Given that $y$ = $s\sqrt{{1-s^2}}+cos^{-1}{s}$, on applying the product rule of the derivative, we get: $\frac{dy}{ds}$ = ${\sqrt{1-s^2}\frac{ds}{ds}\times{s\frac{d{\sqrt{{1-s^2}}}}{ds}}\times-\frac{1}{\sqrt{1-s^2}}}$ or $\frac{dy}{ds}$ = $\sqrt{1-s^2} \times { \frac {-2s^2}{2\sqrt{1-s^2}}\times-\frac{1}{\sqrt{1-s^2}}}$ oso $\frac{dy}{ds}$ = $\frac{1-s^2}{\sqrt{1-s^2}} \times { \frac {-s^2}{\sqrt{1-s^2}}\times-\frac{1}{\sqrt{1-s^2}}}$ The final answer is: = $\frac {-2s^2}{\sqrt{1-s^2}}$