MHB Derivatives of trigonometric functions

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SUMMARY

The derivative of the function y=5tan(x)+4cot(x) is correctly calculated as y' = 5sec^2(x) - 4csc^2(x). The initial claim of y' = -5csc^2(x) is incorrect. The confusion arises from the misunderstanding of the relationship between the two expressions, which are not equivalent identities. The correct derivative aligns with outputs from tools like Wolfram Alpha (W|A).

PREREQUISITES
  • Understanding of basic calculus concepts, specifically derivatives.
  • Familiarity with trigonometric functions such as tangent and cotangent.
  • Knowledge of trigonometric identities and their applications.
  • Experience with mathematical software tools like Wolfram Alpha for verification.
NEXT STEPS
  • Study the rules of differentiation for trigonometric functions.
  • Learn about trigonometric identities and how to prove them.
  • Practice using Wolfram Alpha for calculus problems.
  • Explore advanced calculus topics, such as implicit differentiation.
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Students preparing for calculus exams, educators teaching trigonometric derivatives, and anyone seeking to clarify misconceptions in derivative calculations.

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The answer for derivative of y=5tanx+4cotx is y'=-5cscx^2. But how come on math help the answer is 5sec^2x-4csc^2x? I have a calculus test coming up and I really would appreciate if someone could explain!

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Oh nvm I see my mistake!
 
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Hello, and welcome to MHB! (Wave)

The second answer you cited agrees with what W|A outputs as the derivative for the given function. If the two answers are equivalent, then the following will be an identity (I am assuming you mean something a bit different from the first answer):

$$-5\csc^2(x)=5\sec^2(x)-4\csc^2(x)$$

$$-\csc^2(x)=5\sec^2(x)$$

This is not an identity, so the first result you posted isn't correct. Where did this first result come from?
 

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