Derivative of the inverse of sin(1/x)

In summary: So, I'm not sure why you were told that the answer is different. It may be a typo or a mistake in the given answer. Both answers seem to be equal, as can be seen by simplifying them. Therefore, it seems that both answers are correct.
  • #1
Silversonic
130
1

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.
 
Physics news on Phys.org
  • #2
Silversonic said:

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.
 

FAQ: Derivative of the inverse of sin(1/x)

What is the derivative of the inverse of sin(1/x)?

The derivative of the inverse of sin(1/x) is calculated using the chain rule and is equal to -x^2/(|x| * sqrt(1-x^2)).

Why is the derivative of the inverse of sin(1/x) important?

The derivative of the inverse of sin(1/x) is important because it helps us find the rate of change of a function that involves the inverse of sin(1/x). This can be useful in applications such as physics and engineering.

How do you find the derivative of the inverse of sin(1/x)?

To find the derivative of the inverse of sin(1/x), first rewrite the function as sin^-1(1/x). Then, use the chain rule, which states that the derivative of f^-1(x) is equal to 1/f'(f^-1(x)). In this case, f(x) = sin(1/x), so f'(x) = -cos(1/x) * (-1/x^2). Substituting these values into the chain rule formula gives us the derivative of the inverse of sin(1/x).

What is the domain of the inverse of sin(1/x)?

The domain of the inverse of sin(1/x) is all real numbers except for 0, since sin(1/x) is undefined at x = 0.

Can the derivative of the inverse of sin(1/x) be simplified?

Yes, the derivative of the inverse of sin(1/x) can be simplified to -x^2/(|x| * sqrt(1-x^2)) by factoring out a -1/x^2 from the expression -x^2 * cos(1/x).

Back
Top