Many common functions, such as sin(x), do not have inverse functions because they fail the horizontal line test. (H.L.T.) (Graph y=sin(x) on [-Pi,Pi] to see...)

One way to get around this difficulty is to find a region where the function does satisfy the H.L.T. Such an interval is [-Pi/2, Pi/2] for the sin() function. The inverse of the sin() function is called the inverse sin function, denoted by sin^-1(x). Its domain is [-1,1] and its range is [-Pi/2,Pi/2].

The inverse sin() function is continuous.

Remember

y = sin^-1(x) iff x = sin(y), and

sin(sin^-1(x)) = x = sin^-1(sin(x))

To calculate the derivative of inverse sin(), we use implicit differentiation:

y = sin^-1(x) => x = sin(y) => 1 = cos(y) dy/dx =>

dy/dx = 1/cos(y) = 1/sqrt(1-x²)

Clearly, the chain rule is

d/dx sin^-1(u(x)) = 1/sqrt(1-u²(x)) du/dx

We have, after similar arguments

d/dx ( cos^-1(x) ) = -1/sqrt(1-x²)

Does it make sense that d/dx ( cos^-1(x) ) = -d/dx sin^-1(x) ?

Note: The composition of trig and inverse trig functions requires the use of a triangle...

Note: The remaining derivatives of the inverse trig functions can all be computed quite easily using implicit differentiation!

Last modified Wed Oct 23 23:19:00 CDT 1996