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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

    cristo

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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

    HallsofIvy

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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.

    f'(x)=10*z'(x)
    z'(x)=5*(1+v(x))4*v'(x)
    v'(x)=3*(1-u(x))2*(-u'(x))
    u'(x)=9*7*d(x4)/dx

    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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