### Video Transcript

Find the derivative of the inverse
sine function with respect to π₯.

Weβre differentiating the inverse
sin of π₯ with respect to π₯. So weβre going to begin by letting
π¦ be equal to the inverse sin of π₯. Then we can say that π₯ must be
equal to sin of π¦. Weβre going to differentiate both
sides of this equation with respect to π₯. So we say that d by dπ₯ of π₯ is
equal to d by dπ₯ of sin of π¦. Well, the derivative of π₯ with
respect to π₯ is quite straight forward; itβs one. But weβre going to need to use
implicit differentiation which is a special case of the chain rule to differentiate
sin π¦ with respect to π₯.

The derivative of sin π¦ with
respect to π¦ is cos π¦. So the derivative of sin π¦ with
respect to π₯ is cos π¦ times the derivative of π¦ with respect to π₯ which is just
dπ¦ by dπ₯. So we currently see that one is
equal to cos of π¦ times dπ¦ by dπ₯. We divide both sides of this
equation by cos π¦ to form an equation for the derivative. And we see that dπ¦ by dπ₯ is equal
to one over cos π¦. Now, weβve got a bit of a
problem. We do want an expression for the
derivative in terms of π₯ not π¦.

And remember, we said that π₯ was
equal to sin of π¦. So weβll use the identity cos
squared π plus sin squared π equals one. And Iβve replaced π with π¦. Weβll subtract sin squared π¦ from
both sides of the equation. And then weβll take the square root
of both sides. And we see that cos of π¦ is equal
to the positive and negative square root of one minus sin squared π¦. Remember, inverse sin is restricted
to the closed interval negative π by two to π by two.

By our definition, that means π¦
must be greater than or equal to π two and less than or equal to π by two which,
in turn, means cos of π¦ must be greater than or equal to zero and less than or
equal to one. And thatβs because in the interval
π¦ is greater than or equal to negative π by two and less than or equal to π by
two. The smallest value cos of π¦ weβll
take is zero. And the greatest value is one. And what this means here is weβre
going to take the positive square root of one minus sin squared π¦ only.

We can now replace sin of π¦ with
π₯. And we see that cos of π¦ is equal
to the square root of one minus π₯ squared. And, therefore, dπ¦ by dπ₯ equals
one over the square root of one minus π₯ squared. And we found the derivative of the
inverse sin of π₯. Itβs one over the square root of
one minus π₯ squared for values of π₯ in the range π₯ is greater than negative one
and less than one.