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Creating A Derivative Problem that has a specific solution

  1. Oct 23, 2014 #1
    • OP has received a warning about posting a question with no effort shown
    1. The problem statement, all variables and given/known data
    So this is a problem that I am at a complete loss with. The question asked is, give an equation, using the quotient or power rule that derivative is equal to either sec(x) or cot(x). It doesn't matter which one, sec(x) or cot(x), just as long as the initial equation's derivative is one of them

    Thank you for the help
    Papa
     
  2. jcsd
  3. Oct 23, 2014 #2

    HallsofIvy

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    My first reaction was to use the integral or "anti-derivative" but I suspect this is an exercise preliminary to introducing the anti-derivative.

    You should know the chain rule for differentiation and that the derivative of cos(x) is -sin(x) so you want to find a function, f(u), so that f'(u)= 1/u and then replace u with cos(x).
     
  4. Oct 23, 2014 #3
    Thank you for the quick reply!

    That may be the case, but up until this point and in none of the notes provided for this unit has taught me even the slightest about integrals or the chain rule, could you possibly explain?

    Thanks!
     
  5. Oct 23, 2014 #4

    RUber

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    So, you have learned the quotient rule and the product rule. And I assume you have learned the derivatives of the basic trig functions.
    Your goal is to have a derivative that looks like ##\frac 1 {\cos x} ## or ##\frac {\cos x }{\sin x} ##.
    Have you already learned about the derivative of ##\ln x ##?
    The chain rule says if you have functions of functions i.e. ##f(g(x))## then ##\frac d{dx}f(g(x))=f'(g(x))g'(x)##.
     
  6. Oct 23, 2014 #5
    oooh, alright the chain rule makes sense like that. Thank you very much!
     
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