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Differentiating compound functions

  1. Oct 22, 2005 #1
    I have been happilly solving away a multitude of different questions until the book threw me this curve ball...10^(3x)

    My first attempt was as follows: let y=u^3 and u=10^x
    dy/du = 3u^2...du/d10 = x(10^(x-1))...3x(10^2x(10^(x-1)))...3x(10^(3x-1))

    the answer given in the book however is (3ln10)10^3x...thing is I haven't met a question of this type (they have been of the sort..((x^3)^1/2)/ln(x-2) etc...) and so my best attempt to reach this answer so far is to say that 10^3X is equivilant to saying e^3xln10.

    If y = e^u and u = 3xln10 then...
    dy/du = e^u and du/dx = 3ln10 + (3x/10) giving...(3ln10+(3x/10))10^3x problem is...I have not reached the answer and I am not sure how I've gone wrong...please help!
     
  2. jcsd
  3. Oct 22, 2005 #2

    HallsofIvy

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    I assume that you know that the derivative of ex is just ex itself. You should also have seen (although fewer people feel a need to memorize it!) that the derivative of ax, where a is any positive number, is axln(a).

    If you haven't learned that, then, whenever you have a variable as an exponent, try taking the logarithm of both sides:
    If y= 103x, then ln(y)= 3x ln(10). Now differentiate both sides of that equation, remembering that the derivative of ln(x) is 1/x so the derivative of ln(y) with respect to x is
    [tex]\frac{dln(y)}{dx}= \frac{1}{y}\frac{dy}{dx}[/tex]
    and then solve for [itex]\frac{dy}{dx}[/itex].

    "If u = 3xln10 , the derivative of u is NOT 3ln10+ 3x/10!! "10" is not a variable. u is just ax where a= 3ln10."
     
    Last edited: Oct 22, 2005
  4. Oct 22, 2005 #3
    Thanks for your reply HallsofIvy
    don't remember having seen this(am away from my text book to use the computer...If I have encountered it though I certainly haven't answered any questions that involved it.)I shall check the book later.

    The bit you wrote at the bottom makes sense

    if dy/ydx = 3ln10...dy/dx = (3ln(10))10^3x :smile:

    (for what it's worth I have never come across this technique and so double thanks.)
     
    Last edited: Oct 22, 2005
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