Need help on y=x^tan(x)

  1. if y=x^TAN(x)

    is y'=TAN(x)·x^(TAN(x) - 1)·(SEC(x))^2

    i got the answer by using the chain rule & the power rule by letting u=TAN(X) and y=x^u. not sure if thats the right answer since when i graph the function of y and y' , they don't show any relation with each other.

    i think i have to use the chain rule twice if i want to use the power rule, since x^u does not equal to ux^(u-1) (i think, not exactly sure).

    can someone plz help me out, thanx.
     
  2. jcsd
  3. If

    [tex]y(x)=x^{\tan(x)}[/tex]

    then I get the following result for the derivative:

    [tex]y'(x)=x^{\tan(x)-1}\left(x\log(x)(\sec(x))^2+\tan(x)\right)[/tex]

    or, if you prefer,

    [tex]y'(x)=x^{\tan(x)}\left(\log(x)(\sec(x))^2+\frac{\tan(x)}{x}\right)[/tex]


    I am not quite sure what you mean with:
    but if you are talking about the derivative, then the statement is true:

    [tex]\frac{\partial}{\partial x} x^u = u x^{u-1} \frac{\partial u}{\partial x}[/tex].

    Cheers,
    Freek Suyver.
     
  4. My answer is the same as that of suyver.

    It isn't correct. You cannot let u = (tan x) and then treat it as a constant, because y is a function of x and (tan x) isn't a constant. So we shouldn't apply the power rule directly. Instead, we should first take ln(natural log) on both sides, and then do differentiation.
    The first few steps should be:

    [tex] y = x^{tan x}[/tex]

    [tex]ln y = (tan x)(ln x)[/tex] {take log on both sides}

    Then you can apply the chain rule to finish the rest of the question.
     
  5. thanx.
     
  6. We can do it as suyver has done actually he has done the problem with partial differentiation
     
  7. HallsofIvy

    HallsofIvy 40,385
    Staff Emeritus
    Science Advisor

    In general, if one has y= f(x)g(x), in which both base and exponent are functions of x, one can make either of two mistakes:

    1. Treat the exponent, g(x), as a constant and use the power rule
    y'= g(x)f(x)g(x)-1

    2. Treat the base, f(x), as a constant and use the exponential rule
    y'= ln(f(x))f(x)g(x)


    The interesting thing is that the correct derivative is the sum of these two mistakes!

    y'= g(x)f(x)g(x)-1+ ln(f(x))f(x)g(x)

    as one can show by differentiating ln(y)= g(x)ln(f(x)).
     
    Last edited: Dec 13, 2003
  8. my teacher taught me to use LN rather than LOG

    so i ended up with [x^tan(x)]*(tan(x)/x + sec^2(x)ln(x))
     
  9. HallsofIvy

    HallsofIvy 40,385
    Staff Emeritus
    Science Advisor

    Since there is no good reason to use logarithm to base 10 in higher mathematics (it's used in arithmetic because it works nicely with base 10 numeration), most higher mathematics texts use "log" to mean natural logarithm.
     
  10. are you responding to me? if so, is my equation the equivalent to suyvers?
     
  11. Yes, your equation is equivalent to mine. I always write log(x) for the e-based logarithm and 10log(x) for the 10-based logarithm.
     
  12. HallsofIvy

    HallsofIvy 40,385
    Staff Emeritus
    Science Advisor

    Now that's really strange! Most people use log10(x)!

    hodeez: Yes, I was responding to you. Sorry, I should have quoted your post.
     
  13. "Benifits" of a Dutch highschool education: nlog(x) is the n-based logarithm; log(x) is the 10-based and ln(x) is e-based. At university the e-based logarithm becomes log(x) and people stop using ln(x), but still nlog(x) remains the n-based logarithm. Oh well, just as long as we understand eachother and agree with the final result! :smile:
     
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