Derivative of sin^-1(x) on Interval [1,-1] with Solution Attempt

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SUMMARY

The derivative of the function f(x) = sin-1(x) on the interval [1, -1] is computed as f'(x) = 1/sqrt(1 - x2). To incorporate the interval of π, implicit differentiation is utilized, leading to dy/dx = 1/cos(sin-1(x)), which can be further simplified. Understanding the relationship between the derivative and the specified interval is crucial for accurate computation.

PREREQUISITES
  • Understanding of inverse trigonometric functions, specifically sin-1(x).
  • Knowledge of implicit differentiation techniques.
  • Familiarity with the derivative rules and trigonometric identities.
  • Basic understanding of the interval notation and its implications in calculus.
NEXT STEPS
  • Study implicit differentiation in depth to enhance problem-solving skills.
  • Learn about trigonometric identities related to sin-1(x) and cos(sin-1(x)).
  • Explore the application of derivatives in real-world scenarios, particularly in physics and engineering.
  • Investigate the properties of inverse functions and their derivatives.
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives of inverse trigonometric functions, as well as educators seeking to clarify concepts related to implicit differentiation and interval notation.

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Homework Statement



Compute the derivative of the following function.

Homework Equations



f:[1,-1] arrow [-pie/2, pie/2] given by f(x)=sin^-1 (x)

The Attempt at a Solution



I know that f ' (x)=1/[sqrt(1-x^2)]

Im not sure how to include the intervals of pie given, not sure what they want me to do.
 
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Knowing what the derivative is doesn't do you much good if you have compute it.

Do you know about implicit differentiation?

If so, letting y = f(x), you have y = sin-1(x)
Solve this equation for x, and then calculate dy/dx.

When you do this, you should get dy/dx = 1/cos(sin-1(x)), which you can simplify further. That's where the interval [-pi, pi] comes into play.

BTW, the name of the Greek letter [itex]\pi[/itex] is pi, not pie.
 

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