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Homework Help: Chain rule!

  1. Apr 3, 2009 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    Answers by Microsoft Math 3.0

    3. The attempt at a solution

    This is confusing. Too many parentheses. We used to solve a composite of two or three functions.
    Last edited by a moderator: Nov 29, 2013
  2. jcsd
  3. Apr 3, 2009 #2
    If f(x) = h(g(x)) then f'(x) = h'(g(x)) × g'(x).
    In words: differentiate the ‘outside’ function, and then multiply by the derivative of the
    ‘inside’ function.

    Let's take an example,
    y= sin (x2)
    dy/dx= d(sin x2)/dx *d(x2)/dx

    Treat all the parentheses in a similar manner.
    Last edited: Apr 3, 2009
  4. Apr 3, 2009 #3
    I know, I could solve the third one myself, but the first one :(
  5. Apr 3, 2009 #4
    Break up the expression as:
    z(x) = (1+v(x))5
    v(x) = (2-u(x))3
    u(x) = (6+7x2)9

    finally you have y = 10*z(x)

    now dy/dx = dy/dz *dz/dx
    Last edited: Apr 3, 2009
  6. Apr 3, 2009 #5
    (fog(x) )'=g' (x) f' (g(x) )

    we have this general rule for two functions only. Give me sth for n functions.
  7. Apr 3, 2009 #6


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    shramana has given you a pretty good hint. How about you try to follow it? Post your work, and we will point out any errors.
  8. Apr 3, 2009 #7
    Now let's say
    g(x)= g(h(x))
    so you'll have g'(x)=h'(x)*g'(h(x)) [using f'(x)=g' (x) f' (g(x) )]

    substitute the value of g(x) in f'(x) and so on......
  9. Apr 3, 2009 #8


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    Do it step by step: If y= 10u5, the dy/dx= 50u4 du/dx.

    And u= 1+v3 so du/dx= 3v^2 dv/dx.

    v= ....?
  10. Apr 4, 2009 #9
    OK. Here I've tried to solve the first one.
    Last edited by a moderator: Nov 29, 2013
  11. Apr 4, 2009 #10
    Although it's too long. I won't be able to do it without a software (for multiplication and things like that). Also, We can't use calculator.
  12. Apr 4, 2009 #11
    And the second one which compare to the answer given by the Microsoft Math is wrong:
    Last edited by a moderator: Nov 29, 2013
  13. Apr 4, 2009 #12
    Not at all. You just have to differentiate the functions in the parentheses as you move inwards.


    Now substitute the values of u'(x), v'(x),z'(x) and u(x), v(x) in f'(x).
  14. Apr 4, 2009 #13
    Step 2 is wrong.
    v'(x) = d(ln(ln sec x))/dx
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