MHB Using Chain rule to find derivatives....

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SUMMARY

The discussion focuses on using the chain rule to find the derivative of the function y = (csc(x) + cot(x))^-1. The participants emphasize the importance of defining u as csc(x) + cot(x) and applying the chain rule formula dy/dx = (dy/du)(du/dx). Additionally, they suggest rewriting u in terms of sine and cosine to facilitate the calculation of du/dx, recommending the use of the quotient rule for simplification.

PREREQUISITES
  • Understanding of trigonometric functions: cosecant (csc) and cotangent (cot)
  • Familiarity with the chain rule in calculus
  • Knowledge of the quotient rule for derivatives
  • Ability to manipulate trigonometric identities
NEXT STEPS
  • Practice applying the chain rule with different trigonometric functions
  • Learn about the quotient rule and its applications in calculus
  • Explore trigonometric identities and their derivatives
  • Study advanced derivative techniques, including implicit differentiation
USEFUL FOR

Students and educators in calculus, mathematicians focusing on derivatives, and anyone seeking to deepen their understanding of trigonometric derivatives using the chain rule.

darkknight1
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y = (csc(x) + cot(x) )^-1
Find dy/dx
 
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Suppose u = csc(x) + cot x

dy/dx = dy/du × du/dx
 
darkknight said:
y = (csc(x) + cot(x) )^-1
Find dy/dx
darkknight, you titled this "using chain rule" and Monoxdifly just told you what that is! Are you able to do this now?

Letting u= csc(x)+ cot(x), y= u^-1. What is dy/du? What is du/dx?
chain rule: dy/dx= (dy/du)(du/dx).

If the difficulty is du/dx, it might help you to write [math]u= csc(x)+ cot(x)= \frac{1}{sin(x)}+ \frac{cos(x)}{sin(x)}= \frac{1+ cos(x)}{sin(x)}[/math] and use the "quotient rule".
 

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