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Differentiation check

  1. Aug 18, 2009 #1
    1. The problem statement, all variables and given/known data

    By using cos and sin subs for tan and sec, find the gradient of:

    ln(tan2x+secx)

    2. Relevant equations

    tanx=sinx/cosx

    secx=1/cosx

    3. The attempt at a solution

    Substituting

    [tex]y=ln(\frac{sin2x}{cos2x}+\frac{1}{cosx})[/tex]

    Using the chain rule let:

    [tex]z= \frac{sin2x}{cos2x}+\frac{1}{cosx}[/tex]


    [tex]y=lnz[/tex]

    [tex]\frac{dy}{dz}=\frac{1}{z}=z^{-1}=(\frac{sin2x}{cos2x}+\frac{1}{cosx})^{-1}=\frac{cos2x}{sin2x}+cosx[/tex]


    looking at z use the quotient rulee for the first part sin2x/cos2x let:

    [tex]u=sin2x[/tex] [tex]\frac{du}{dx}=2cos2x[/tex]

    [tex]v=cos2x[/tex] [tex]\frac{dv}{dx}=-2sin2x[/tex]

    [tex]\frac{dz}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex]

    [tex]=\frac{(cos2x)(2cos2x)-(sin2x)(-2sin2x)}{(cos^22x)}[/tex]

    [tex]=\frac{(2cos^22x)+(2sin^22x)}{(cos^22x)}[/tex]

    [tex]=\frac{2(cos^22x)+(sin^22x)}{(cos^22x)}[/tex]

    As [tex]sin^2x+cos^2x=1[/tex] therefore:

    [tex]=\frac{2}{(cos^22x)}[/tex]


    Going back to the now partionally differentated value of z

    [tex]y=\frac{1}{cosx}=cos^{-1}x[/tex]

    if [tex]y=cos^{-1}x[/tex] then [tex]x=cosx[/tex]

    so [tex]\frac{dx}{dy}=-siny[/tex] so [tex]\frac{dy}{dx}=\frac{-1}{sin y}[/tex]

    since - [tex]cos^2y+sin^2y=1[/tex] then [tex]sin^2y=1-cos^2y=1-x^2[/tex]

    meaning [tex]siny=\sqrt{1-x^2}[/tex] and [tex]\frac{d}{dx}{cos^{-1}}=\frac{-1}{\sqrt{1-x^2}}[/tex]

    That menas that[tex]\frac{dz}{dx}=\frac{2}{(cos^22x)}-\frac{1}{\sqrt{1-x^2}}[/tex]


    Back to the fist chain rule:

    [tex]\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}=(\frac{cos2x}{sin2x}+cosx).(\frac{2}{(cos^22x)}-\frac{1}{\sqrt{1-x^2}})[/tex]



    Sorry for the mass workings but I've been struggling with this and have now had a brain failure and can't see if I am right or even how to simplify the final differential.

    Is what I have done correct?
     
  2. jcsd
  3. Aug 18, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No! 1/(a+ b) is NOT equal to (1/a) + (1/b).


     
  4. Aug 18, 2009 #3
    Thanks for the input. Have I started off using the method by using the chain rule?

    Do I just leave it as:

    [tex](\frac{sin2x}{cos2x}+\frac{1}{cosx})^{-1}[/tex]

    ?
     
  5. Aug 18, 2009 #4
    You have another mistake:
    [tex]\frac{1}{cosx} \neq cos^{-1}x = arccosx[/tex]
    You can look here on how to differentiate something like 1/f(x) here
    http://en.wikipedia.org/wiki/Reciprocal_rule
    Look at the last example on the page :wink:

    One more note: it would have been much easier to differentiate tan2x using u=2x than changing it to sin2x/cos2x, although you did get that part correct.

    That, or put it in the denominator as /(tan2x + secx), or find a common denominator and add them, then take the reciprocal to get rid of the -1.
     
  6. Aug 22, 2009 #5
    I'm still not 100% with this question.

    I have been re-working on it and have now come up with this answer.

    [tex]\frac{dy}{dx}=\frac{\frac{2(sin2x)^2}{(cos2x)^2}+\frac{sinx}{(cosx)^2}+2}{\frac{sin2x}{cos2x}+\frac{1}{cosx}}[/tex]

    Does this look right?
     
  7. Aug 24, 2009 #6
    Where did that last 2 come from in the numerator at the end? If you take that out, you have the correct derivative of ln(tan2x + secx).
     
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