Answer
$$y' = -cos(x)\sqrt{1+sin^2x}$$
Work Step by Step
$$y=\int\limits_{sinx}^1 {\sqrt{1 + t^2}}dt$$
Switch the bounds, making the integral negative
$$y=-\int\limits^{sinx}_1 {\sqrt{1 + t^2}}dt$$
Substitute the upper bound in for t and multiply by the derivative.
$$y' = -\sqrt{1 + sin^2(x)} \times (sinx)'$$
Simplify.
$$y' = -\sqrt{1 + sin^2(x)} \times (cosx)$$
$$y' = -cos(x)\sqrt{1+sin^2x}$$