Welcome to a lesson on
the derivatives of inverse
trigonometric functions.
Let's go and take a look
at our derivative rules.
When you first look at this,
it may be overwhelming.
But upon closer inspection,
we should notice
that the right column is just the negative
of the left column.
So that should help us remember these.
Again the derivative of arc sine u,
and the derivative of arc
cosine u are just opposites
of one another.
And the same is true for the derivative
of arc tangent u and arc cotangent u.
And the derivative of arc secant u
and arc cosecant u.
Let's see if we can derive
at least one of these,
before we can do some examples.
Remember that the arc sine function
is the inverse of the sine function.
If we start with y equals sine x
and want to find the inverse function.
The procedure is to interchange
the x and the y variables
and then solve for y.
Let's go ahead and just interchange the x
and the y variables.
So here we have the inverse sine function,
where y would be the angle and x would be
the sine function value.
Let's go ahead and model
that in this right triangle.
Our angle would be y and
the sine function value
would be x or x over one.
So the ratio of the opposite
side of the hypotenuse
would be x to one.
Now using the Pythagorean theorem,
we could find the length of this side
as the square root of one minus x squared.
Okay, let's go back to
our inverse function
and let's find the derivative of this,
and see if we can get it to
match this derivative formula.
So we'll take the derivative of both sides
with respects to x.
You can see we'll have to
use implicit differentiation.
The derivative of x would equal one.
The derivative of sine
y with respects to x
would be cosine y times dy dx.
To solve for dy dx, we divide
both sides by cosine y.
Let's go ahead and rearrange this.
We have dy dx equals
one divided by cosine y.
If we go back to our triangle,
the cosine of angle y
is equal to the ratio
of the adjacent side to the hypotenuse,
for the square root of one minus x squared
divided by one.
So this would equal one
divided by the square root
of one minus x squared.
And this would be the derivative of
the inverse sine function.
You take a look at what we just did
and the formula for the
derivative of arc sine u,
they are exactly the same, except this one
is in terms of u, implying
we may have to use
the chain rule.
In this one we didn't have
to use the chain rule.
But now you can see where this
derivative rule came from,
and we can verify the other
five in a very similar fashion.
But let's go and take a
look at a few examples.
We want to determine the
derivative of arc tangent
square root x.
Here's our derivative rule.
Notice u would equal the square root of x.
Or x to the power of 1/2.
We do have to find u
prime, let's do that now.
U prime would be 1/2
times x to the power of
1/2 minus one or negative 1/2.
Well this is the same as 1/2
times the square root of x.
Move this down, the
denominator and divided
as the square root of x.
Now we can apply our differentiation rule.
F prime of x is equal to u prime,
1/2 square root of x, divided
by one plus u squared.
Well the squared of x
squared would just be x.
This is our derivative,
but we cannot leave it
in this form, remember a fraction bar
is just a division symbol.
So this will be 1/2 square root of x.
Instead of dividing by one plus x,
we could multiply by the reciprocal,
which would be one over one plus x.
So our derivative would
be one divided by two
square root of x, times
the quantity one plus x.
Let's take a look at another one.
Here we have f of x
equals arc secant five x.
So u is equal to five x.
So u prime is equal to five.
So f prime of x is equal to u prime,
which will be five, divided
by the absolute value
of u, absolute value of five x,
times the square root
of u squared minus one.
U squared would be 25 x squared minus one.
Now we can simplify this
a little bit further.
This is a positive five inside
the absolute value symbol.
If we factor this five
out, it's still going to be
positive, therefore we can
simplify this factor five
with this factor five.
Notice the x still stays
inside the absolute value.
So we would have one divided
by the square root of x,
times the square root of
25 x squared minus one.
Now don't be tempted to simplify this,
this is a difference of squares
where we're factoring the five x plus one
and the five x minus one.
So we don't have two equal factors here.
So this is as simplified as we can get it.
So now we have f of x equals
two times the arc cosine
of x over three.
So u is equal to x divided by three,
or we can think of this as 1/3 x.
So u prime equals 1/3.
Here's our derivative rule.
So f prime of x equals negative u prime,
so negative 1/3, divided
by the square root
of one minus u squared.
One minus u squared with the
x squared over three squared
or x squared over nine.
Again this is our derivative but it's not
in very good form.
Let's see if we can simplify this.
Remember this is just a division problem.
So this is negative 1/3
times the reciprocal
of our denominator.
So it would be one over
square root one minus x
squared over nine.
And let's go ahead and multiply
these fractions together.
The product is a negative,
the numerator is equal to one.
Now let's look at our denominator.
This three is outside the square root.
If we were to bring it
inside the square root,
it would change to a nine,
because the square root
of nine would give us
this three odds here.
So this is really the
same as the square root
of nine times one minus x
squared divided by nine.
And the reason we might want to do this,
is because now we can
eliminate the fractions
underneath the square root.
This would simplify to negative one over
square root of nine minus x squared.
And this is typically how most textbooks
prefer the answer to be written.
And you can see why, it does
look a bit more simplified.
So the final form of the
derivative would be this.
I think we have time for one more.
Determine the derivative of f of t equals
sine of arc cosine t.
Let's see if we can simplify this function
before we try to find its derivative.
Arc cosine t would
return some angle theta.
Where the cosine function value would
be equal to t or t over one.
So if we let this angle equal theta,
the ratio of the adjacent
side of the hypotenuse
would be t to one.
Now using the Pythagorean
theorem we could show
that the length of this
side would have to be
square root of one minus t squared.
Let's go back to our function now.
Arc cosine t returns some angle theta,
and now we want to find
the sine of angle theta.
Well the sine of angle theta would be
the ratio of the opposite
side of the hypotenuse,
or the square root of one
minus t squared to one.
Which means that this function f of t
is really just the sine of that angle,
which would be the square
root of one minus t squared.
So now we can find the
derivative of this function,
rather than the derivative
of this composite
function involving an
inverse trig function.
To do that we'd probably
want to rewrite this as
f of t equals the quantity
one minus t squared
to the 1/2 power.
So to find the derivative
of this we'll apply
the power rule with the chain rule.
So the derivative of the outer function
would be 1/2, there's our
u to the 1/2 minus one,
that'd be negative 1/2, times u prime.
Well this is our u, so
it'd be negative two t.
Now we'll simplify.
This two and this two simplify.
And so our derivative
is equal to negative,
t would be in the
numerator, and this would be
in the denominator, so
we'd have the square root
of one minus t squared.
Okay, I hope you found
these explanations helpful.
Thank you for watching.
