Unizor - Derivatives Example - arcsin(x)
Channel: Zor Shekhtman on UNIZOR Education
Date: 13/12/2016

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Derivative Example -
Inverse Trigonometric Functions

f(x) = arcsin(x)

f I(x) = 1/√1−x²


Proof

We will use the method of implicit differentiation to obtain the formula for a derivative of this function.

Let's start from a definition of function arcsin(x).

The domain of this function is [−1,1] and its values are in [−π/2,π/2].

Then, for any value of the argument x from its domain the function value y is defined as an angle in radians such that
(a) sin(y)=x
(b) −π/2 ≤ y ≤ π/2

The first statement can be expressed as
sin(arcsin(x))=x
Since these two functions, A(x)=sin(arcsin(x)) and B(x)=x, are equal within domain [−1,1], their derivatives are equal as well.

The derivative of the A(x) can be obtained using the chain rule for compounded functions.
AI(x)=d/dx[sin(arcsin(x))] = cos(arcsin(x))·d/dx[arcsin(x)]

The derivative of B(x) is trivial.
BI(x) = d/dx[x] = 1

From equality of these two derivatives we conclude
d/dx[arcsin(x)]=1/cos(arcsin(x))

Now let's analyze the expression cos(arcsin(x)).
We know that
sin(arcsin(x))=x and
arcsin(x)∈[−π/2,π/2].
Therefore,
cos(arcsin(x)) ≥ 0.
Hence,
cos(arcsin(x)) = √1−x²

The final formula for a derivative is
[arcsin(x)]I = 1/√1−x²

A small detail remains when x = ±1, which results in zero denominator. These are the points where our function arcsin(x) is not differentiable.
Geometrically, it signifies that tangential lines at both ends, x=−1 and x=1, of the domain of function arcsin(x) are vertical, as can be seen from a graph of this function below: