# Derivative of the inverse of sin(1/x)

1. Jan 17, 2012

### Silversonic

1. The problem statement, all variables and given/known data

Find the derivative of the compositional inverse of $f(x) = sin(1/x)$ restricted to (1,∞). You may use without proof that sin(x) is differentiable with derivative cos(x).

2. Relevant equations

$(f^{-1})'(y_0) = \frac{1}{f'(f^{-1}(y_0))}$

3. The attempt at a solution

The compositional inverse of $f(x) = sin(1/x)$ is $f^{-1}(y_0) = \frac{1}{arcsin(y_0)}$.

Plugging that in to the equation gives;

$(f^{-1})'(y_0) = \frac{1}{f'(\frac{1}{arcsin(y_0)})} = \frac{1}{sin'(arcsin(y_0))} = \frac{1}{sin'(1/x)} = \frac{1}{(-1/x^2)cos(1/x)} = \frac{-x^2}{cos(1/x)}$

And by putting back in

$x = \frac{1}{arcsin(y_0)}$

$(f^{-1})'(y_0) = \frac{-(\frac{1}{arcsin(y_0)})^2}{cos(arcsin(y_0))} = - \frac{1}{\sqrt{1-y_0^2}arcsin^2(y_0)}$

However, I'm told that the answer is;

$\frac {1}{cos(\frac{1}{arcsin(y_0)})}$

For the life me, I can't see to get the answer given? Even wolfram alpha confirms that what I have it correct.

http://www.wolframalpha.com/input/?i=derivative+of+1/arcsin(y)

Are we both correct? Because I can't see to show they are both equal to each other.

2. Jan 17, 2012

### SammyS

Staff Emeritus
I get the same result you get.