Differentiation (Trig Function)

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SUMMARY

The discussion focuses on taking the second derivative of the function cos(x^2). The first derivative is correctly calculated as -2xsin(x^2). To find the second derivative, participants suggest using the product rule and chain rule, leading to the expression 2x(-2xcos(x^2) - 2sin(x^2)). The final confirmation of the solution is sought, indicating a collaborative effort to ensure accuracy in differentiation techniques.

PREREQUISITES
  • Understanding of basic differentiation rules, including the product rule and chain rule.
  • Familiarity with trigonometric functions and their derivatives.
  • Knowledge of composite functions and how to differentiate them.
  • Ability to manipulate and simplify algebraic expressions in calculus.
NEXT STEPS
  • Study the application of the product rule in differentiation.
  • Learn about the chain rule in detail, especially with composite functions.
  • Practice differentiating trigonometric functions involving polynomial arguments.
  • Explore examples of higher-order derivatives for complex functions.
USEFUL FOR

Students studying calculus, particularly those focusing on differentiation techniques, as well as educators looking for examples of collaborative problem-solving in mathematics.

SciGuy26
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Homework Equations



Take the second derivative of:
cos(x^2)

The Attempt at a Solution



cos(x^2)

dy/du = -sin(u)
du/dx = 2x

dy/dx= -2xsin(x^2)

I don't know how to begin taking the derivative of this a second time...
 
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What rule do you think you will need to use?
 
Use the product rule with -2x and sin(x^2). You'll have to use the chain rule again when you calculate the derivative of sin(x^2).
 
So I'd have 2xcos(u) + -2sin(u)

How am I to use the chain rule at this point?

Ugh I don't understand how combine rules with composite functions.
 
nm...

Solution should be as follows:
-2xsin(x^2)

dy/du= sin(u)
du/dx= 2x

-2xsin(u) + -2xsin(U)

-2xcos(u) + -2sin(u)

= 2x(-2xcos(x^2) + -2sin(x^2))

Could someone please confirm? Thank you for the help.
 

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