### Video Transcript

In this video, weβll learn how to
find the derivatives of the inverses of trigonometric functions. We will learn how to do this by
using implicit differentiation. And thus, it is important that you
understand how to apply the chain rule, if not have a thorough understanding of how
implicit differentiation works, before watching this video. After deriving the derivatives of
the inverse trigonometric functions, weβll consider the application of these
derivatives to more complicated inverse trigonometric functions.

Before we look at finding the
derivative of our inverse trigonometric functions, letβs just quickly consider the
function π¦ equal sin of π₯. Remember, π₯ is a real number. And of course, since weβre
performing calculus with a trigonometric function, we need to make sure this is
measured in radians. For this function, we can say that
π₯ is equal to the inverse sin of π¦. This subscript negative one denotes
the inverse function. Remember though without restricting
the domain of arc sin of π₯ or inverse sin of π₯, the function is going to be many
to one. We, therefore, restrict the domain
for π of π₯ equals inverse sin of π₯. And we say that π₯ has to be
greater than or equal to negative one and less than or equal to one.

If we go back to the function π₯
equals the inverse sin of π¦, we can, therefore, see that π₯ will take values
greater than or equal to negative π by two and less than or equal to π by two. Weβll also need to recall the chain
rule. This says that if π¦ is some
function in π’ and π’ is some differentiable function in π₯, then dπ¦ by dπ₯ is
equal to dπ¦ by dπ’ times dπ’ by dπ₯. Weβll now use everything weβve seen
here to find the derivative of the inverse sine function.

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 [negative] π by 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.

In our next example, weβll consider
an alternative method that will help us find the derivative of the inverse cosine
function.

This time, weβre going to need to
know the inverse function theorem. This says that if π is a
differentiable function with a continuous inverse π prime and π prime of π is not
equal zero, then not only is π invertible but it has a differentiable inverse. Such that the derivative of the
inverse of π at some π equals π of π is equal to one over the derivative of π
at π. This is sometimes written simply as
dπ₯ by dπ¦ is equal to one over dπ¦ by dπ₯. Letβs see how this might help us
when differentiating the inverse cosine function.

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.

In our next example, weβll consider
how we might apply the process thatβs used so far to find the derivative of the
inverse tangent function.

Find an expression for the
derivative of π¦ equals the inverse tan of ππ₯ in terms of π₯.

Since π¦ is equal to the
inverse tan of ππ₯, we can write ππ₯ as being equal to tan π¦. Weβre going to use implicit
differentiation to find the derivative of both sides of this equation. The derivative of ππ₯ with
respect to π₯ is simply π. And the derivative of tan π¦
with respect to π₯ is equal to the derivative of tan π¦ with respect to π¦ times
the derivative of π¦ with respect to π₯. The derivative of tan π₯ is sec
squared π₯. And the derivative of π¦ with
respect to π₯ is dπ¦ by dπ₯.

So we see that π is equal to
sec squared π¦ times dπ¦ by dπ₯. Dividing through by sec squared
π¦ and we see that dπ¦ by dπ₯ equals π over sec squared π¦. Weβre going to need to
represent our equation for the derivative in terms of π₯. So weβll use this trigonometric
identity. One plus tan squared π₯ equals
sec squared π₯. This means we can write dπ¦ by
dπ₯ as π over one plus tan squared π¦. And then we replaced tan π¦
with ππ₯. And we see the expression for
the derivative of π¦ equals the inverse tan of ππ₯ is π over one plus ππ₯
squared.

Similar rules can be applied to
help us find the derivative of the inverse cotangent function. We find that the derivative of the
inverse of cotangent of π of π₯ is equal to negative π over one plus ππ₯
squared. The inverse cosecant and secant
functions are a little more unusual. So weβll consider next how to find
the derivative of the inverse cosecant function.

Find d by dπ₯ of the inverse
cosecant of π₯.

We begin by letting π¦ be equal
to the inverse cosecant of π₯. And this means we can rewrite
this. And we can say that π₯ is equal
to the cosecant of π¦.

Weβre next going to use
implicit differentiation to find the derivative of both sides of this
equation. The derivative of π₯ with
respect to π₯ is simply one. Then the derivative of cosec π¦
with respect to π₯ is equal to the derivative of cosec π¦ with respect to π¦
times dπ¦ by dπ₯. And the derivative of cosec π¦
with respect to π¦ is negative cosec π¦ cot π¦. So we see that one is equal to
negative cosec π¦ cot π¦ times dπ¦ by dπ₯.

Now, we know that, for the
inverse cosecant function, π¦ must be greater than negative π by two and less
than π by two and not equal to zero. Using these restrictions cosec
π¦ cot π¦ cannot be equal to zero. So we can divide through by
negative cosec π¦ cot π¦. And we see that dπ¦ by dπ₯ is
as shown. We want to represent our
equation for the derivative in terms of π₯. So weβll use this trigonometric
identity cot squared π¦ plus one equals cosec squared π¦. And we can rewrite this to say
that cot of π¦ is equal to the positive and negative square root of cosec
squared π¦ minus one.

We put this into the equation
for the derivative in place of cot of π¦. And we then use the fact that
π₯ is equal to cosec π¦. But we are going to need to
make a decision on the sine of the derivative. And it can help here to look at
the graph of the inverse cosecant function. Notice how, for all values of
π₯ in the range of the function, the derivative of the slope of the tangent is
negative.

And we, therefore, use the
absolute value to ensure that our derivative is always negative. We say that dπ¦ by dπ₯ is equal
to the negative of the absolute value of one over π₯ times the square root of π₯
squared minus one. Since one and the square root
of π₯ squared minus one are always positive, we can rewrite this as shown. So the derivative of the
inverse cosecant function π₯ with respect to π₯ is negative one over the modulus
or absolute value of π₯ times the square root of π₯ squared minus one.

A similar process can be applied to
help us find the derivative of the inverse secant function. And we have the derivatives of all
of the inverse trigonometric functions we require. Itβs useful to commit these results
to memory but also be prepared to derive them where necessary. Weβll now have a look at the
application of these results.

Evaluate the derivative of the
inverse cotangent of one over π₯ with respect to π₯.

Here, we have a function of a
function or a composite function. Weβre, therefore, going to need
to use the chain rule to find the derivative. This says that if π and π are
differentiable functions such that π¦ is π of π’ and π’ is π of π₯, then dπ¦
by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. Weβll let π’ be equal to one
over π₯. Then π¦ is equal to the inverse
cot of π’. To apply the chain rule, we
need to find the derivative of both of these functions. And with π’ it can be useful to
write it as π₯ to the negative one.

Then dπ’ by dπ₯ is negative π₯
to negative two or negative one over π₯ squared. We can then use the general
derivative of the inverse cotangent function. And we see that dπ¦ by dπ’ is
equal to negative one over one plus π’ squared. dπ¦ by dπ₯ is the product of
these. Itβs negative one over π₯
squared times negative one over one plus π’ squared.

We can replace π’ with one over
π₯ and then multiply through. And we see that the derivative
of the inverse cotangent of one over π₯ with respect to π₯ is one over π₯
squared plus one.

Did you notice that the derivative
of the inverse cotangent of one over π₯ is equal to the derivative of the tangent of
π₯? This is, in fact, no accident. And we can use the identity the
inverse cotangent of one over π₯ equals the inverse tan of π₯. That could have saved us a little
bit more time in this previous example.

Evaluate the derivative of the
inverse sin of the square root of one minus π₯ squared with respect to π₯.

Here, we have a function of a
function or a composite function. So weβll use the chain rule to
find its derivative. This says that if π¦ is some
function in π’ and π’ is some function in π₯, then dπ¦ by dπ₯ is equal to dπ¦ by
dπ’ times dπ’ by dπ₯. Weβll let π’ be equal to the
square root of one minus π₯ squared. Which can, of course,
alternatively be written as one minus π₯ squared to the power of one-half. Then π¦ is equal to the inverse
sin of π’. To apply the chain rule, weβre
going to need to find the derivative of both of these functions. The derivative of the inverse
sin of π’ with respect to π’ is one over the square root of one minus π’
squared.

And we can use the general
power rule to find the derivative of one minus π₯ squared to the power of
one-half. Itβs a half times one minus π₯
squared to the negative one-half times the derivative of the bit inside the
brackets which is negative two π₯. That can be written as negative
π₯ times one minus π₯ squared to the power of negative one-half.

dπ¦ by dπ₯ is, therefore,
negative π₯ over the square root of one minus π₯ squared times one over the
square root of one minus π’ squared. We can replace π’ with one
minus π₯ squared to the power of one-half. And the second fraction becomes
one over the square root of one minus one minus π₯ squared. This further simplifies to one
over π₯. And we divide through by
π₯. And we see that the derivative
of our function is negative one over the square root of one minus π₯
squared.

Once again, weβve stumbled across
an interesting result. That is that the derivative of the
inverse sin of the square root of one minus π₯ squared is equal to the derivative of
the inverse cos of π₯. This comes from the identity of the
inverse sin of the square root of one minus π₯ squared is equal to the inverse cos
of π₯, the values of π₯ between zero and one. Being familiar with this result
could have reduced the amount of what we needed to do in this example.

In this video, weβve seen that we
can use implicit differentiation or the inverse function theorem to derive the
formulae for the derivatives of inverse trigonometric functions. We saw that the derivatives of the
inverse trigonometric functions are as shown. And we also saw that being familiar
with certain trigonometric identities can sometimes significantly reduce the process
of finding these derivatives.