Derivative of the inverse of sin(1/x)

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SUMMARY

The discussion focuses on finding the derivative of the inverse function of f(x) = sin(1/x) restricted to the interval (1, ∞). The correct derivative is established as (f^{-1})'(y_0) = 1/cos(1/arcsin(y_0)), while the user's calculation yields -1/(sqrt(1 - y_0^2) * arcsin^2(y_0)). Both results are confirmed by Wolfram Alpha, leading to confusion regarding their equivalence. The key takeaway is the importance of correctly applying the inverse function derivative formula and understanding the relationship between the sine and arcsine functions.

PREREQUISITES
  • Understanding of inverse functions and their derivatives.
  • Familiarity with the sine and arcsine functions.
  • Knowledge of differentiation rules, particularly for composite functions.
  • Basic proficiency in calculus, specifically regarding limits and continuity.
NEXT STEPS
  • Study the properties of inverse trigonometric functions.
  • Learn about the application of the chain rule in differentiation.
  • Explore the relationship between sine and arcsine through graphical representation.
  • Investigate common pitfalls in calculating derivatives of inverse functions.
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Students studying calculus, particularly those focusing on derivatives of inverse functions, as well as educators seeking to clarify concepts related to trigonometric identities and their derivatives.

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Homework Statement



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).

Homework Equations



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

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 into 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.
 
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Silversonic said:

Homework Statement



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).

Homework Equations



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

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 into 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)}

\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)}

However, I'm told that the answer is;

\displaystyle\frac {1}{cos(\displaystyle\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.
I get the same result you get.
 

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