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Back to The Chain Rule

Exercise 2

Avg difficulty 2.5

Consider the function k(θ)=sin⁡−1(sin⁡(θ))\fe{k}{\theta}=\fe{\sin^{-1}}{\fe{\sin}{\theta}}.

Over the interval [−π2,π2]\cinterval{-\frac{\pi}{2}}{\frac{\pi}{2}}, k(θ)=θ\fe{k}{\theta}=\theta. What does this tell you about the formula for k′(θ)\fe{\fd{k}}{\theta} over (−π2,π2)\ointerval{-\frac{\pi}{2}}{\frac{\pi}{2}}?

Simonds, Calculus Lab Manual, §6.5 Supplement, Exercise 2

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Adapted from Calculus Lab Manual © 2012–2016 Portland Community College. Licensed under CC BY-SA 4.0. Source: Calculus Lab Manual. Imported at revision 3060f8e.