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Derivative of the natural log

  1. Dec 22, 2006 #1
    1. The problem statement, all variables and given/known data
    find dy/dx for y=ln(sec^(2)x)

    2. Relevant equations

    3. The attempt at a solution
    1.)1/(sec(x)^2)*dy/dx sec(x)^2


    3.)i put it in terms of sin and cos...

    4.)i cancelled the cos(x) in the denominator and came out with sin(x)cos(x)
    the answer i am supposed to get is 2tan(x). where did i go wrong?
  2. jcsd
  3. Dec 22, 2006 #2
    The derivative of (sec(x))^2 is NOT tan(x).

    Edit: Btw, 1/(sec(x)^2)*dy/dx sec(x)^2, should actually be (1/(sec(x)^2))*d(sec(x)^2)/dx
  4. Dec 22, 2006 #3
    Solving by chain rule, you get-:

    d(ln(sec^2(x)))/d(sec^2(x)) * d(sec^2(x))/d(sec x) * d(sec x)/d(x)

    Solving finally, you get 2tanx
  5. Dec 22, 2006 #4
    oh, right sorry about that on the edit.
    and what is the derivative of (sec(x))^2, because i know that the derivative of tan(x) is (sec(x))^2 so i assumed it would work the other way around
  6. Dec 22, 2006 #5
    no,the other way back is by integrating the derivative to get the original function
    derivative of (sec x)^2 is 2secx.secxtanx
  7. Dec 22, 2006 #6
    oh ok i get it now, thanx for your help. easy mistake
  8. Dec 22, 2006 #7


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    Notational fix:

    1.)1/(sec(x)^2)*dy/dx sec(x)^2

    This is wrong. What you meant to say was something like

    [tex]\frac{1}{\sec^2 x} \cdot \frac{d}{dx}\left( \sec^2 x \right)[/tex]


    [tex]\frac{1}{\sec^2 x} \cdot \frac{d(\sec^2 x)}{dx}[/tex]


    [tex]\frac{1}{\sec^2 x} \cdot \frac{du}{dx}
    \quad \quad (u = \sec^2 x)[/tex]

    (In particular, you did not mean to put a y in there. You've already defined [itex]y=\ln \sec^2 x[/itex], and that's certainly not what you wanted in this particular expression)
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