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A handy trick i found

  1. Oct 28, 2005 #1
    I'm guessing not a lot will care about this becasue it's not very relevant, but my calc teacher couldn't do this and I did it in a few seconds, so i'll expose it.

    The problem stated that f(x)=x^3+x

    and inverse of f(x)=g(x) and g(2)=1

    question: Find g'(2)

    ----------------------------

    My teacher tried to create a formula to connect inverse derivative answers and inverse functions for cubics. He couldn't. So while staring at it I realize how the derivative is dy/dx which is appearing everywhere you derivate a y.

    so I write y=x^3+x
    take inverse x=Y^3+y
    and I don't care about what the function looks like. I don't worry about putting it in standard form like he tried. I keep it like this and take derivative. Out of nowhere I might say I had written down 1=3y^2 dy/dx + dy/dx

    and isolating the dy/dx => dy/dx = 1/(1+2y^2)

    since point (2,1) was given, the fact that I have no x is not important. i can plug in y instead. And I get the final answer. g'(2)=1/4

    The relevance of this is that finding the derivative of a function can be expressed in many forms, related to various letters in that expression. many times the y' has both x and y.

    But...Standard form was not important here, and pretty much everyone, myslef included for a few minutes were hooked up on putting it in standard form...

    I thought i'd share this with you.

    ~Robokapp
     
  2. jcsd
  3. Oct 28, 2005 #2
    Very impressive, I wish I had that sort of intuitive math skill. See If you can extend that particular problem to a more genralized formula or set of steps. If you can find some overarching shortcut, you could save yourself a whole lot of time. Try additional polynomial terms and there inverses. Tell us what happens
    -see....we care-
     
  4. Oct 28, 2005 #3
    but i don't think i understand. how is dy/dx [ y^3 ] = 3y^2? isn't it zero?
     
  5. Oct 28, 2005 #4
    Substituting x into the first equation gives:

    [tex](y^3+y)^3 + y^3+y\ne y[/itex]

    So how is that the inverse?
     
  6. Oct 28, 2005 #5

    Hurkyl

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    Given the relation y=x³+x, which defines the graph of y=f(x), its inverse relation is x=y³+y, which defines the graph of y=g(x).

    Or, to write it more clearly: (IMHO)

    We have that f(x)=x³+x (for all x in the domain of f), so we also have that x=g(x)³+g(x). (for all x in the domain of g)


    You've discovered and applied one of the powerful methods of differential calculus. Congratulations!


    Now, the next question is to generalize. :smile:

    For any function f, whose inverse is g, can you write down a formula for g'(a)?
     
  7. Oct 28, 2005 #6
    Is that what I would perhaps call implicit differentiation?
     
  8. Oct 28, 2005 #7

    Hurkyl

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    Yep! (Well, one of the major steps in this calculation is called implicit differentiation)
     
  9. Oct 28, 2005 #8
    well, y is a variable, not a constant.

    let's pick y=2x => dy/dx=2 agreed?

    but it is that because dy/dx is the derivative of y. basically you derivate both sides. Everywhere you see a y you add a dy/dx at the end of it because it doesn't contain an x. y' or dy/dx is same thing...
     
  10. Oct 28, 2005 #9
    Silly teacher didn't know implicit differentiation! What level is the original poster at? Congratulations to him/her for working this out for him/herself, though.
     
  11. Oct 28, 2005 #10
    hah. I got it!

    I'm Calculus AB Highschool senior by the way.

    the question: What is the derivative of the inverse of any function f(x)

    the answer:

    1/[f'(y)]

    -----------------------------

    I will post my work. I will quote exactly what I wrote in a wordpad document to generalize this:

     
  12. Oct 28, 2005 #11

    Hurkyl

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    Yep, that's the right formula! :smile:

    Judging from your post, though, you've only proven it for polynomials. Can you figure out how to prove it for an arbitrary differentiable, invertible function? (Hint: it's the same basic idea, but the details are simpler than what you've done with polynomials)
     
  13. Oct 28, 2005 #12
    let's try a simple f(x)=e^x

    my formula says that the derivative of inverse should be 1/f'(y)

    so 1 /[ d/dx of e^(y)]
    1/[e^(y)*(y')] is what I seem to get.

    1/(y'f(y)) maybe!

    ------------

    I can't prove it...I get stuck in a big pile of information andit's too late for that...I'll take a look tomorrow.

    What I'll try to see is is 1/[y'f(y)] same as 1/f'(y)

    tops cancel. y'f(y)=f'(y)

    by Chain Rule the derivative of f(y) is f'(y)*y'

    Oh shoot. I chose a bad example, didn't I? e^x and its derivative are the same so I can't tell if my formula is 1/[y'f(y)] or 1/[y'f'(y)]

    Well, it's friday night and I'm already dizzy so I'll trust my previews work and the Chain Rule and state that the formula 1/[y'f'(y)] works.

    However...it still won't work meaning I'm terribly wrong. I give up...I'll look at it tomorrow! :surprised
     
  14. Oct 28, 2005 #13
    Think about the chain rule.
     
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