Derivative of the inverse of sin(1/x)

[Silversonic]

Homework Statement

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

Homework Equations

[itex] (f^{-1})'(y_0) = \frac{1}{f'(f^{-1}(y_0))}[/itex]

The Attempt at a Solution

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

Plugging that into the equation gives;

[itex] (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)} [/itex]

And by putting back in

[itex] x = \frac{1}{arcsin(y_0)} [/itex]

[itex] (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)} [/itex]



However, I'm told that the answer is;

[itex] \frac {1}{cos(\frac{1}{arcsin(y_0)})} [/itex]



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.

 

[SammyS]

Homework Statement

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

Homework Equations

[itex] (f^{-1})'(y_0) = \frac{1}{f'(f^{-1}(y_0))}[/itex]

The Attempt at a Solution

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

Plugging that into the equation gives;

[itex] (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)} [/itex]

And by putting back in

[itex] x = \frac{1}{arcsin(y_0)} [/itex]

[itex]\displaystyle (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)} [/itex]

However, I'm told that the answer is;

[itex]\displaystyle\frac {1}{cos(\displaystyle\frac{1}{arcsin(y_0)})} [/itex]

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.

I get the same result you get.