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

${\sqrt {1+x^2}}$
Let's take the derivative of both sides! $\frac{dy}{dx}=\frac{d}{dx}\int_{0}^{x} {\sqrt {1+t^2}} dt$ Let's use the fundamental theorem of Calculus (substitute the limits, and multiply by their derivatives)! $\frac{dy}{dx}= {\sqrt {1+x^2}}\cdot 1-{\sqrt {1+0^2}}\cdot 0= {\sqrt {1+x^2}}$