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Finding Derivatives with Respect to a Variable

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data

    The question says ti find the derivative of y with respect to the independent variable.
    The equation is: y=4 ln 2t.


    2. Relevant equations

    I know how to find derivatives using the product/quotient/chain/etc rules, but that isn't what the question is asking for. I though it wanted me to use implicit differentiation, but that doesn't really make sense because y is explicitally defined in terms of x. I don't really know what should be used.

    3. The attempt at a solution

    Before, I tried to just get the derivative normally and got 4 ln 2t +2/t, but that's not right. Then I also tried using implicit differentiation, but it wasn't working.

    By the way, the answer is (ln4/t)4 ln 2t

    Thank you in advance!
     
  2. jcsd
  3. Mar 28, 2012 #2
    What's the derivative of ln(x) ?

    If you know the derivative of that, then you can simply apply the chain rule.

    [tex]\frac{dy}{dx} = 4\frac{d}{dx}ln(2x)[/tex]

    recall that [tex]\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}[/tex]
     
  4. Mar 28, 2012 #3
    The derivative of ln(x) is 1/x.

    Am I supposed to use implicit differentiation, then? Or is the chain rule enough?
     
  5. Mar 28, 2012 #4
    The chain rule is plenty.

    What you can do is simply use a substitution. Let some function U(t) equal the "contents" of the original (ln) function.

    Then find the derivative of ln(U), then apply the chainrule.
     
  6. Mar 28, 2012 #5
    The derivative of ln(U) would just be 1/U then. I would only be working with the equation 4ln(U). After I use the chain rule for that equation, would I substitute the original ln function?
     
  7. Mar 28, 2012 #6
    4 is a constant so it just comes outside, then let f(x) = ln(U), and let U(x) = 2t then simply recall the chain rule:

    [tex]4\frac{df}{dx} = 4\frac{df}{du} \frac{du}{dx}[/tex] //I just multiplied in a 4 there for the constant
     
  8. Mar 28, 2012 #7
    Okay. I get it now. This would work for all problems like this, right?
     
  9. Mar 28, 2012 #8
    Well yes. Technically you ALWAYS are using the chain rule. It's just that for cases like the derivative of y=x (y'=1), the "inner product" or composite function's derivative is just 1 also.

    If you have a ton of functions, let's say something like this:

    y = x, x = t, t = p

    You can see that you could just substitute it all in, or you can simply follow the chain rule:

    [tex]\frac{dy}{dp} = \frac{dy}{dx} \frac{dx}{dt} \frac{dt}{dp} [/tex]

    and so on, up to an infinite number of nestled composite functions.
     
  10. Mar 28, 2012 #9
    Great! This was the one thing I was worried about, but it doesn't seem to be too bad now. Thank you!
     
  11. Mar 28, 2012 #10
    What solution did you get?
     
  12. Mar 28, 2012 #11
    I got (ln4/t)4 ln 2t.
     
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