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Homework Help: What is the derivative of (tan(x))^(1/x) and sqrt((x-1)(x^2 (x-4)))?

  1. Apr 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the following derivatives:

    [tex]\frac{d}{dx}[/tex] (tanx)^(1/x)

    [tex]\frac{d}{dx}[/tex] [tex]\sqrt{\frac{x-1}{x^2 (x-4)}}[/tex]

    2. Relevant equations
    Rules of differentiation

    3. The attempt at a solution
    I realize these are fairly elementary problems, just not sure the order I should perform the differentiation rules. My attempt at number 1 is as follows:

    Considering 1/x as a separate function (u), we can use the chain rule:

    [tex]\frac{d}{dx}[/tex]tan(x)[tex]_{u}[/tex] [tex]\frac{d}{dx}[/tex] [tex]\frac{1}{x}[/tex]

    Apply quotient rule on 1/x:

    = sec[tex]_{2}[/tex] (x)[tex]_{u}[/tex] [tex]\frac{1}{x^2}[/tex]

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

    Can someone please correct me if I have made any mistakes?

    The second problem I'm not sure. As far as I can tell it's just applying the quotient rule to the fraction, all to the power of [tex]\frac{1}{2}[/tex]. Do I need to apply the chain rule to the denominator before using the quotient rule?
  2. jcsd
  3. Apr 16, 2009 #2
    Re: Derivatives...

    Well the only mistake I see in your first work is that you forgot a negative. and for the second problem break it down to a more simple problem. It may become more work to resimplify it later but you aren't wrong to do so.

    You can take it and make it into this d/dx((x-1)(1/2)/(x2(x-4))(1/2)) which can then become d/dx((x-1)(1/2)*(x3-4x2)(-1/2)). Which is much easier to differentiate by the product rule or course you can always stop one step short of this and use the quotient rule.
  4. Apr 16, 2009 #3
    Re: Derivatives...

    To do the first one, you will either need to use logarithmic differentiation, or rewrite the expression as

    e^ln((tanx)^(1/x)) which is e^((1/x)(ln(tanx)))

    Now the base is a constant, in fact e, so it looks like e^u... use the chain rule now
  5. Apr 16, 2009 #4
    Re: Derivatives...

    Missing a negative where?

    You sure about that one? Just seems like a lot of work for such a simple function.
  6. Apr 16, 2009 #5
    Re: Derivatives...

    Consider the following function: x^x

    Find (d/dx)x^x

    You cannot use either of the rules (d/dx)x^n = nx^(n-1) or (d/dx)e^x = e^x.
    In the first rule the exponent is a CONSTANT, in the second the base is a CONSTANT (in particular it is e.)
    You also cannot use the more general version of the second rule, (d/dx)a^x = ln(a)*a^x because the base is a CONSTANT.

    In the expression x^x, neither the base nor the exponent are constant.

    So you use logarithmic differentiation, or you rewrite it as e^(ln(x^x)) which is e^(x*lnx) then, let u = x*lnx and it looks like e^u, at which point you can use the chain rule along with the fact that (d/dx)e^x = e^x.
  7. Apr 16, 2009 #6
    Re: Derivatives...

    Sorry this took so long. Originally I was looking at how you did it not the question and what the answer should be. You forgot a negative from d/dx(1/x) but I'm pretty sure you can't do it that way. As far as I know you can approach this straight on or using a ln trick. I'll show both.

    Straight on
    1)Take derivative of exponent and put at front then subtract 1 from part with exponent then tack on the derivative of the inside on the end. Like so:

    Or with ln

    1) set it equal to y
    2)ln both sides
    3)move (1/x) to front in accordance with laws of ln
    5)times by y on either side and sub back in to eliminate it since y=tan(x)^(1/x)
  8. Apr 16, 2009 #7
    Re: Derivatives...

    Questions about the previous post:

    ??? d/dx(1/x)(tan(x))^((1/x)-1)d/dx(tan(x)) What rule is being used?

    You cannot use d/dx x^n = n*x^(n-1) here (with or without a chain rule.)

    ??? ln(y)=(1/x)ln(tan(x)) ===> dy/y=(-1/x^2)(1/tan(x))(sec(x)^2) You need to use the product rule here.
  9. May 5, 2009 #8
    Re: Derivatives...

    It would be the chain rule but in tandum with the product rule which I forgot to do anyway, and thank you for catching that.
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