Question

find dy/dx using fundamental theorem of cal part1. (i actually have no idea how to do this one...)

Answer 1

\[y = \int\limits_{0}^{\sin x} dt/\sqrt{1-t ^{2}},\left| x \right|<\pi/2\]

Answer 2

\[\int\limits_{0}^{sinx}\frac{ 1 }{ \sqrt{1-t^2} }dt\]
The fundamental theorem only works if the lower limit is a constant, which it is in this case, and a single variable for the upper limit.

So let u = sinx
Keep the chain rule in mind as well is we can no longer use d/dt and instead must use
d/du * du/dt

\[\frac{ d }{ du }\int\limits_{0}^{u}\frac{ 1 }{ \sqrt{1-t^2} }dt \frac{ du }{ dt }\]

The integral can now be differentiated.

\[\frac{ 1 }{ \sqrt{1-u^2} }\frac{ du }{ dt }\]

The derivative of u is cosx since we defined it as sinx earlier.

\[\frac{ cosx }{ \sqrt{1-u^2} }\]

Now just substitute u back in

\[\frac{ cosx }{ \sqrt{1-\sin^2x} }\]

Answer 3

thank you so very much for the simple to understand explanation!