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Homework Help: More chain rule

  1. Jul 31, 2010 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    y'=7(-csc^2)^6(x^5) * 5x^4
  2. jcsd
  3. Jul 31, 2010 #2
    Here, you actually have three functions, so you have to apply the chain rule twice.

    We have [tex] f(g(h(x))), [/tex] where

    [tex] h(x) = x^5 [/tex]

    [tex] g(x) = cot(x) [/tex]

    [tex] f(x) = x^7 [/tex]

    When applying the chain rule, the approach is from "outside to inside," in that we start from the outer most function and systematically go towards the innermost function. Remember, when doing the chain rule, each step is distinct.

    You have: [tex] 7(-csc^2)^6..... [/tex]

    You were correct in taking the derivative of f(x) = x^7 first; however, you need to let everything inside f(x) remain untouched during this step. Meaning, for f(g(h(x)), we leave g(h(x)) untouched.

    So what you should have had is: [tex] 7cot^6(x^5).... [/tex]

    Try to do the rest by yourself.
  4. Jul 31, 2010 #3
    so is it

    7cot^6(x^5) * (-csc^2)(x^5) * 5x^4

    = -35x^4 * cot^6(x^5) * csc^2 * x^5
  5. Jul 31, 2010 #4
  6. Jul 31, 2010 #5
    I still have a question, when you have cot^7 i get leaving it alone to begin with, but then when you take the derivative to get -csc^2, why don't you drop the 7 to get -7(csc^2)^6
  7. Jul 31, 2010 #6
    Because it's

    [tex] [f( g(x) )]' = f'(g(x)) * g'(x), [/tex]

    and not

    [tex] [f( g(x) )]' = f'(g'(x)). [/tex]
  8. Aug 1, 2010 #7


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    In other words, to differentiate [itex]cot^7(x^5)[/itex] you think: "The 'outer' function is a 7th power so first I have [itex]7 cot^6(x^5)[/itex]. Now the 'next' inner function is a cotangent, "cot(x^5)", so I have to multiply that by its derivative: [itex]-csc^2(x^5)[/itex]. Finally, the last function is [itex]x^5[/itex] so that has to be multiplied by its derivative, [itex]5x^4[/itex].

    [tex](cot^7(x^5))'= (7 cot^6(x^5))(-csc^2(x^5))(5x^4)= -35x^4cot^6(x^5)csc(x^5)[/tex]
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