Derivative of Inverse Trig Function

So the problem indicates:

Suppose y= sin x , for x in [n/2, 3n/2] *note let n=pi

Then we define P to be the inverse of this function.

Ex. P(0) = n

a. Obtain the derivative of P. The derivative is not P'(x) = 1/[(1-x^2)^(1/2)].

So I'm not entirely sure how to go about this.

We can write P(x) =y for x in [-1, 1].

Then I know that on the interval [n/2, 3n/2] the derivative of the sine function will be negative.

But I'm unsure how to approach it?

Re: Derivative of Inverse Trig Function

Hi, "turbozz."

I am puzzled.

If y = sin(x), and P is the inverse of this function, then P = x = arcsin(y).

And the range changes: instead of being stated in terms of x, it is stated in terms of y: [1, -1].

By definition, the derivative of the arcsin function is:

Why is the derivative *not* this standard definition?

Are you looking for a derivative at one particular point?

Re: Derivative of Inverse Trig Function

I'm puzzled about this to. But I know it has do with fact y=P(x) is defined for x in [-1,1] and y in [n/2, 3n/2], rather than in the conventional way y=arcsinx is defined where y in [-n/2, n/2]. When I draw the graph of P(x) it does vary from the graph of y= arcsinx (for y in

[-n/2, n/2]), so the slopes must be different => derivatives must be different. But I don't know how to find P'(x).

Re: Derivative of Inverse Trig Function

The difference is that while "normal" inverse sine returns a value between and , this function is defined to return a value between and . One way to handle that is to simply "shift" the returned value. but, as we are told, so we can write . (Notice that we are **subtracting** , not adding it. That is because the sine function, after goes **down** rather than up as it does after . Of course, the derivative of the constant is 0 so the derivative of P(x) is the **negative** of the derivative of .

Re: Derivative of Inverse Trig Function

Thank HallsofIvy. It completely agrees with my graph of P(x).