Answer
$y' = -\sqrt{1+sin^2(x)}\times cos(x)$
Work Step by Step
We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $ y =\int_{sin(x)} ^{1} \sqrt{1+t^2}dt$
Swap upper and lower bounds: (this makes the expression negative)
$ y = -\int_{1} ^{sinx} \sqrt{1+t^2}dt$
Since the upper bound is a function we need to use the chain rule:
$u = sin(x)$
$u' = cos(x)$
$ y =-\int^{u} _{1} \sqrt{1+t^2}dt$
Using the FTC
$F(u) =\int_0^uf(θ)dθ$
$F'(u) = f(u)$
Thus,
$y'= -\sqrt{1+u^2}\times u'$
$y' = -\sqrt{1+sin^2(x)}\times cos(x)$
