# Derivative of Inverse Trig Function

• Nov 16th 2013, 09:29 AM
turbozz
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)].

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?
• Nov 16th 2013, 11:15 AM
DavidB
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:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} y}\arcsin(y) = \frac{1}{\sqrt{(1-y^{2})}}$

Why is the derivative not this standard definition?

Are you looking for a derivative at one particular point?
• Nov 16th 2013, 11:34 AM
turbozz
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).
• Nov 16th 2013, 11:42 AM
HallsofIvy
Re: Derivative of Inverse Trig Function
The difference is that while "normal" inverse sine returns a value between $\displaystyle -\pi/2$ and $\displaystyle \pi/2$, this function is defined to return a value between $\displaystyle \pi/2$ and $\displaystyle 3\pi/2$. One way to handle that is to simply "shift" the returned value. $\displaystyle sin^{-1}(0)= 0$ but, as we are told, $\displaystyle P(0)= \pi$ so we can write $\displaystyle P(x)= \pi- sin^{-1}(x)$. (Notice that we are subtracting $\displaystyle sin^{-1}(x)$, not adding it. That is because the sine function, after $\displaystyle x= \pi/2$ goes down rather than up as it does after $\displaystyle -\pi/4$. Of course, the derivative of the constant $\displaystyle \pi$ is 0 so the derivative of P(x) is the negative of the derivative of $\displaystyle sin^{-1}(x)$.
• Nov 16th 2013, 12:01 PM
turbozz
Re: Derivative of Inverse Trig Function
Thank HallsofIvy. It completely agrees with my graph of P(x).