Results 1 to 8 of 8

Math Help - Derivative of (tan(x))^x ?? :s

  1. #1
    Junior Member
    Joined
    Dec 2008
    From
    London
    Posts
    45

    Derivative of (tan(x))^x ?? :s

    been trying to this for some time now dont get it lol :S

    (tan(x))^x

    i tried chain rule :s

    let y=u^x and u=tan(x)

    y'=ln(u)/u , u'=sec^2(x)

    \frac{dy}{dx}=\frac{sec^2(x).ln|tan(x)|}{tan(x)}

    but ans is [ln|tan(x)|+2xcosec(2x)](tan(x))^x
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    The standard method is to take natural logs of both sides and then differentiate...

    However, - Wolfram|Alpha does it with the chain rule for two inner functions...

    \frac{d}{dx}\ f(u(x), v(x)) = \frac{\partial f}{\partial u} \frac{du}{dx} + \frac{\partial f}{\partial v} \frac{dv}{dx}

    ... which doesn't mean we have to, of course, but I fancy this picture makes it palatable...



    ... where



    ... is the double version of...



    ... the ordinary chain rule. As with that, straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed lines similarly but with respect to the (corresponding) dashed balloon expression which is (one of) the inner function(s) of the composite expression.

    Shading shows that an expression has been treated as a constant on the way down (during differentiation).

    You do, admittedly, need logarithmic differentiation to see the logic of the double-dashed differentiation, i.e. the logic of...

    \frac{d}{du}\ a^u = \ln a\ a^u


    _____________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus: Standard Integrals, Derivatives and Methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; November 26th 2009 at 10:27 AM. Reason: pic
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2008
    From
    London
    Posts
    45
    I kinda get the jist of what you posted but can you clarify what the lines are getting confused by your solid, dashed, dobule dashed lines lol :S
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Sure - travelling down a dashed line means differentiating with respect to the dashed balloon, so that other expressions are held as constants. So it's all just equivalent to the two-variable chain rule - which I'll fill out in a minute. Meanwhile, why not click the Wolfram link and click there on 'show steps'...

    \frac{d}{dx}\ f(u(x), v(x)) = \frac{\partial f}{\partial u} \frac{du}{dx} + \frac{\partial f}{\partial v} \frac{dv}{dx}

    So...

    \frac{d}{dx}\ (\tan x)^x = \frac{\partial}{\partial (\tan x)}\ (\tan x)^x \frac{d}{dx}\ \tan x + \frac{\partial}{\partial x}\ (\tan x)^x \frac{d}{dx}\ x

    Notation here possibly not entirely kosher, but I hope it helps.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Dec 2008
    From
    London
    Posts
    45
    Alright cool thanks i understood your way and i also tried the differentiation by logs method and both gave me the following answer.

    \frac{dy}{dx}=(tan(x))^x[(x(sec(x)csc(x)))+ln|tan(x)|]

    but i don't see how in the back of the book they got it as this form

    <br />
[ln|tan(x)|+2xcosec(2x)](tan(x))^x<br />

    someone explain somehow they simplified

    (x(sec(x)csc(x))) to 2xcsc(2x)

    thankkks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    sin(2x) = 2 sin x cos x
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Dec 2008
    From
    London
    Posts
    45
    mmmm not sure :X

    trying this

    csc(x)sec(x)=1/sin(x)cos(x)

    using 1/2sin(2x)=sin(x)cos(x)

    csc(x)sec(x)=1/[1/2sin2x] = 2sin2x = 2[2cosxsinx] but i dont c me getting newhere near 2csc2x from there on
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Quote Originally Posted by Kevlar View Post
    mmmm not sure :X

    trying this

    csc(x)sec(x)=1/(sin(x)cos(x))

    using 1/2sin(2x)=sin(x)cos(x)

    csc(x)sec(x)=1/[(1/2)sin2x] = 2/sin(2x) ...
    Nearly there
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. contuous weak derivative $\Rightarrow$ classic derivative ?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 22nd 2011, 02:37 AM
  2. Replies: 0
    Last Post: January 24th 2011, 11:40 AM
  3. [SOLVED] Definition of Derivative/Alt. form of the derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 23rd 2010, 06:33 AM
  4. Derivative of arctan in a partial derivative
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 12th 2010, 01:52 PM
  5. Replies: 2
    Last Post: November 6th 2009, 02:51 PM

Search Tags


/mathhelpforum @mathhelpforum