Maximizing Efficiency: Applying the Chain Rule in Differentiating AC Sources

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SUMMARY

The discussion focuses on applying the chain rule in differentiating the function v = 4sin^3(4t). The solution involves breaking down the function using trigonometric identities and applying both the product and chain rules effectively. The final derivative calculated is dv/dt = 8 cos(4t) + 16 sin(4t) sin(8t) - 8 cos(4t) cos(8t). Participants emphasize the importance of correctly identifying the inner function for proper application of the chain rule.

PREREQUISITES
  • Understanding of trigonometric functions and identities
  • Familiarity with the chain rule in calculus
  • Knowledge of product and quotient rules in differentiation
  • Basic skills in manipulating algebraic expressions
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  • Study the application of the chain rule in more complex functions
  • Explore advanced trigonometric identities and their derivatives
  • Learn about the product and quotient rules in differentiation
  • Practice solving differential equations involving trigonometric functions
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Students studying calculus, particularly those focusing on differentiation techniques, as well as educators looking for examples of applying the chain rule with trigonometric functions.

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Homework Statement



Chain/product or quotient

Homework Equations



v=4sin^3 (4t)


The Attempt at a Solution



NB:@ Means teta(spelling might be wrong)

4sin^3 (4t) = 4sin (4t).sin (4t).sin (4t)

using sin^2@=1-cos 2@ /2


v= 4sin (4t). (1-cos 2@ /2)


= 4sin (4t)/2 . 4sin (4t). cos (8t)/2

= 2 sin (4t) - 2 sin (4t) cos (8t)




then I done some calculation containing product and chain rule with a final answer of


dv/dt = 8 cos (4t) + 16 sin (4t) sin (8t) - 8 cos (4t) cos (8t)



Thanx for your help
 
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hi maali5!
maali5 said:
v= 4sin (4t). (1-cos 2@ /2)

no, v= 4sin (4t). (1-cos 2@)/2

but anyway you should have used the chain rule, with g(t) = sin(4t) :wink:
 

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