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Homework Help: Differentiation help

  1. Apr 7, 2008 #1
    Help please on these. I'm extremely rusty with differentiation and want to know if I've got the right answers here
    1. The problem statement, all variables and given/known data
    differentiate the following functions (you do not have to simplify):
    a. f(x)=[tex](x-2)^{3}e^{-2x}[/tex]
    b. f(x) = cos x / ln ([tex]x^{2}[/tex] + x)
    c. g(t) = [tex]cos^{4}[/tex]([tex]t^{2} + e^{2t}[/tex])
    d. f(x) = ln (h(x))
    e. [tex]y = (sin x)^{ln x}[/tex] x>0

    3. The attempt at a solution
    First three I think I've got:
    a. I used the product rule and came out with the following:

    b. -([tex]x^{2}[/tex] + x) sin x / (2x + 1)

    c. [tex]-8(t + e^{2t})cos^{3}(t^{2} + e^{2t})sin(t^{2} + e^{2t})[/tex]
    d and e I'm stuck on! Should I use the Chain rule for d and e?

    If a, b and/or c are wrong, any pointers would be awesome. cheers.
  2. jcsd
  3. Apr 7, 2008 #2
    Let's start with a): It's not quite right. How did you get your answer?
  4. Apr 7, 2008 #3
    Well, all of your answers are wrong. This is because while applying the product/quotient rule, you also need to apply the chain rule. Not sure what you did for c). Try rechecking.
  5. Apr 7, 2008 #4
    Well I did say I was very rusty!
    okay, looking at them again I think where I'm going wrong is that I just differentiating parts seperately, so not using the Product rule correctly. Is that fair to say?

    For a), Is this closer to the answer:
    [tex](x-2)^{3}.-2e^{-2x} + 3(x-2)^{2}.e^{-2x}[/tex]
    and then combine like terms from there?
  6. Apr 7, 2008 #5
    Correct, now simplify as you said!
  7. Apr 8, 2008 #6


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    (d) is probably the easiest of the problems. Since you don't know what "h(x)" is, you will need to leave the answer in terms of h'(x).

    for (e), and generally when you have the variable in both exponent and base, use "logarithmic differentiation". That is, let y= (sin(x))ln(x) and take the logarithm of both sides: ln(y)= ln(x) ln(sin(x)). Now differentiate both sides (using the chain rule extensively) and solve for y'.
  8. Apr 10, 2008 #7
    Right, I think I'm getting on okay with these. Thanks for all the help.
    I'm still having issues with c though.
    I tried doing it a different way:
    Now correct me if I'm wrong but:
    [tex]cos^{4}(t^{2} + e^{2t})[/tex]
    can be written as
    [tex](cos(t^{2} + e^{2t}))^{4}[/tex]
    right? (please say it is!)

    And then using the chain rule we can do the following:
    [tex]4cos^{3}(t^{2} + e^{2t}).-sin(t^{2} + e^{2t}).(2t + 2e^{2t})[/tex]
    simplifying from there, I end up with the same answer as per my OP, which Snazzy said was wrong. I'm more inclined to believe Snazzy at the mo than my own (very limited) calc abilities. So any pointers as to where I'm going wrong?

    For (d), for the answer do I make the outside function is the logarithm and the inside is h(x)?
    So the answer is:
    f'(x) = h'(x)/h(x)
  9. Apr 10, 2008 #8
    Your answer for c is right. It was I who probably made the mistake, or didn't bother to check the third answer.

    Your answer for d is right.
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