### Video Transcript

Find the derivative of the
inverse cos of π₯ over π with respect to π₯, where π is not equal to zero.

Weβll begin by letting π¦ be
equal to the inverse cos of π₯ over π. This can be alternately written
as π₯ over π equals cos of π¦. And we can then multiply both
sides by π. And we see that π₯ is equal to
π times cos of π¦. Weβre going to differentiate
our expression for π₯ with respect to π¦. In other words, weβre going to
find dπ₯ by dπ¦. Weβll use the general result
that the derivative of cos of π₯ with respect to π₯ is negative sin of π₯. And we see that dπ₯ by dπ¦ must
be equal to negative π sin of π¦.

Now, before we perform the next
step we need to recall the fact that for the inverse trigonometric functions, we
restrict their domains. And we know that the domain of
the inverse cos of π₯ or our cos of π₯ is greater than or equal to zero and less
than or equal to π. This means that π¦ must be
greater than or equal to zero and less than or equal to π. Now, weβre going to use the
inverse function theorem. So weβre going to take values
of π¦ greater than zero and less than π, such that sin of π¦ is not equal to
zero.

Using this criteria, we can use
dπ₯ by dπ¦ equals one over dπ¦ by dπ₯ which can be rearranged to say that dπ¦ by
dπ₯ equals one over dπ₯ by dπ¦. And we see that, for our case,
dπ¦ by dπ₯ equals one over negative π sin of π¦. And we have an expression for
the derivative in terms of π¦. Remember, we want this to be in
terms of π₯. We said that π₯ over π equals
cos of π¦. So weβll use the fact that sin
squared π¦ plus cos squared π¦ equals one and rearrange this to say that sin π¦
equals plus or minus the square root of one minus cos squared π¦. When π¦ is between zero and π,
sin π¦ is greater than zero. So in fact, weβre only
interested in the positive root.

So weβll replace this in our
expression for the derivative. And we get negative one over π
times the square root of one minus cos squared π¦. We then replace cos π¦ with π₯
over π and change π₯ over π all squared to π₯ squared over π squared. And then we bring π into the
square root. And we see that dπ¦ by dπ₯ is
equal to negative one over the square root of π squared minus π₯ squared. So d by dπ₯ of the inverse cos
of π₯ over π is equal to negative one over the square root of π squared minus
π₯ squared for values of π₯ between negative π and π.