What Causes Undefined Derivatives in Trigonometric Functions?

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SUMMARY

The discussion centers on finding the relative extrema of the function abs(sin(2x)) over the interval 0 < x < 2π. The derivative is identified as 2 cos(2x) for x > 0, but it is noted that the derivative does not exist at x = π/2, x = π, and x = (3π)/2. This is due to the absolute value function affecting the continuity and differentiability at these points. The key takeaway is that the sign of sin(2x) is critical in determining where the derivative is defined.

PREREQUISITES
  • Understanding of trigonometric functions, specifically sine and cosine.
  • Knowledge of calculus concepts, particularly derivatives and extrema.
  • Familiarity with absolute value functions and their impact on differentiability.
  • Ability to analyze graphs of functions to identify critical points.
NEXT STEPS
  • Study the properties of absolute value functions in calculus.
  • Learn how to find critical points and relative extrema in piecewise functions.
  • Explore the implications of continuity and differentiability in trigonometric functions.
  • Practice graphing functions like |sin(2x)| to visualize changes in behavior at critical points.
USEFUL FOR

Students and educators in calculus, mathematicians focusing on trigonometric analysis, and anyone interested in understanding the behavior of derivatives in piecewise-defined functions.

Nick_OCD
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Hello All,

My problem is to find the relative extrema of abs(sin 2x); 0< x < 2pi.

I got the derivative right, I think: 2 cos 2x if x > 0, and -2 cos 2x if x < 0.

However, the solution says that f ' (x) does not exist at x = pi/2, x = pi, and x = (3pi)/2.

Glancing at my trigonometry charts, I see that sin and cos are never undefined,

so why does 2 cos 2x or -2 cos 2x not exist at pi/2, pi, or 3pi/2?

Thanks,

Nick
 
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Nick_OCD said:
Hello All,

My problem is to find the relative extrema of abs(sin 2x); 0< x < 2pi.

I got the derivative right, I think: 2 cos 2x if x > 0, and -2 cos 2x if x < 0.

But your interval only includes x > 0. And it isn't the sign of x that is important, it is the sign of sin(x) because of the absolute value signs.

However, the solution says that f ' (x) does not exist at x = pi/2, x = pi, and x = (3pi)/2.

Glancing at my trigonometry charts, I see that sin and cos are never undefined,

so why does 2 cos 2x or -2 cos 2x not exist at pi/2, pi, or 3pi/2?

Thanks,

Nick

Draw the graph of |sin(2x)| and you will answer your own questions.
 
Thank you. I had a brain freeze on that one.
 

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