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Derivative of Inverse

  1. Nov 18, 2004 #1
    I was taught in class that if g(x) is f^-1(x), then g'(y)=1/f'(x)

    my notes for an example:
    f(x)=x^2
    g(x)=f^-1(x)=x^(.5)
    g’(4)=?
    g(x)= x^(.5)
    g’(x)=.5x^-.5
    g’(4)=1/f’(x), x=4^(.5)
    g’(4)=1/f'(2)=1/2*2=1/4

    I have no clue how g(x)=f^-1(x)=x^(.5) and how g’(4)=1/f'(2)

    I am just overall confused any help would be appreciated
     
  2. jcsd
  3. Nov 18, 2004 #2
    f(x) = x^2

    set f(x) = y

    y = x^2

    (*To find inverse solve for x explicitly in terms of y*)

    y^0.5 = x

    f^-1(x) = x^0.5

    let g(x) = f^-1(x)

    g(x) = x^0.5

    The second one follows the same way
     
  4. Nov 18, 2004 #3
    I have:
    g’(4)=1/f'(2)=1/2*2=1/4

    and also:
    g'(4)=1/f'(x)=1/f'(2)=1/4

    is both ways acceptable?
     
  5. Nov 18, 2004 #4
    I wouldn't go with the second one. It could be seen as wrong to say g'(4) = 1/f'(x) without saying what x is explicitly, which in this case is root(4).

    But, it depends how picky your instructor is
     
  6. Nov 18, 2004 #5
    what about:
    g’(x)=.5x^-.5=.5(4)^-.5=1/4

    could that be acceptable too?
     
  7. Nov 18, 2004 #6
    Sure,

    g(x) = [tex]x^\frac{1}{2}[/tex]

    thus,

    g'(x) = [tex]\frac{1}{2x^\frac{1}{2}}[/tex]

    g'(4) = [tex]\frac{1}{2*4^\frac{1}{2}}[/tex]

    g'(4) = [tex]\frac{1}{2*2}[/tex]

    g'(4) = [tex]\frac{1}{4}[/tex]


    Again, I would just recommend that you specify you are changing from the general equation g'(x) to g'(4) somewhere in your solution. But if its not on a test, or your teacher isn't picky then that should be fine.
     
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