Derivatives of the Inverse Trigonometric Functions

DRAFT

From the derivative rules we know that

Remember that these formulas apply only when x is measured in radians.

But what about their inverses, sin−1(x), cos−1(x) and tan−1(x)?

[otherwise denoted by arcsin(x), arcos(x) and arctan(x)]

There is a simple trick that allows us to find their derivatives:

The derivative of sin−1(x)

Let y = sin−1(x)  x = sin(y)

Now, instead of differentiating y with respect to x, we differentiate x with respect to y

(in other words, we reverse the usual roles of x and y).

Differentiate x = sin(y) with respect to y



Now rearrange the trigonometric identity sin2(y) + cos2(y) = 1 to express cos(y) in terms of sin(y):

sin2(y) + cos2(y) = 1

 cos2(y) = 1 - sin2(y)

 cos(y) =

But x = sin(y), so cos(y) =

Substitute this into and we get

Finally take the reciprocals of both sides:



But, hang on a minute. Take a look at the denominator of

For to be defined 1 − x2 > 0  −1 < x < 1

So here we have the answer:

The derivative of cos−1(x)

The derivative of cos−1(x) can be found in exactly the same way. You can try to do this one.

and the result is:

The derivative of tan−1(x)

This follows a similar method, but is a little different:

Let y = tan−1(x)  x = tan(y)

Differentiate x = tan(y) with respect to y



Now rearrange the trigonometric identity tan2(y) + 1 = sec2(y) to express sec2(y) in terms of tan2(y):

tan2(y) + 1 = sec2(y)

 sec2(y) = 1 + tan2(y)

But x = tan(y), so sec2(y) = 1 + x2

Substitute this into and we get

Finally take the reciprocals of both sides:



For to be defined, 1 + x2  0 which is true for all real values of x.

So here we have the answer:

Integration

The derivatives of the trigonometric functions can also be expressed as integrals.

Remember from the Introduction to Integrals that ...

... finding an Integral is the reverse of finding a Derivative.

Therefore:

Remember that we need to add a constant of integration because the derivative of a constant is zero.

Why are the derivatives of sin−1(x) and cos−1(x) so similar?

The answer to this question is simple. It is because sin−1(x) + cos−1(x) is always a multiple of

Don’t believe me?

Try a few values:

1. Let’s try x = where sin−1(x) and cos−1(x) are angles in the first quadrant.

Remember that the angles must be measured in radians.

and

So

2. Let’s try x = -0.8 where sin−1(x) and cos−1(x) are angles in the third quadrant.

and

So 4.0688... + 3.7850... = 7.8539... =

So, if sin−1(x) + cos−1(x) =

it follows that = 0

 =

Therefore, or

where c1 and c2 are constants that differ by a multiple of

Because there are two alternative answers for you will usually see the version only, but both are correct.

Differential equations

Certain types of first order differential equations have solutions involving the inverse trigonometric functions.

Example

Solve the differential equation

by the homogeneous method

, so is of the form

Therefore

Let y = vx, so

Therefore



Separate the variables:









Now substitute back y = vx

