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Homework Help: Finding derivatives of inverse trig functions using logarithms

  1. Apr 16, 2013 #1
    For some polynomial functions it is useful to logarithmize both sides of the eq. First. How can this be applied for inverse trig functions? Is it even possible?
  2. jcsd
  3. Apr 16, 2013 #2


    Staff: Mentor

    I can't see how this would be useful for inverse trig functions, or even how it would be useful for polynomials. Polynomials generally consist of a sum of terms, and there is no property that lets you simplify the log of a sum.
  4. Apr 16, 2013 #3
    Sorry for that. I meant product and quotient of polynomial expressions.

    Now About the inverse trig function: I saw in a book a list of exercises where they said to apply that logarithmization technique and there were some inverse trig functions, but I it was a mistake: They probably mixed topics...
  5. Apr 16, 2013 #4


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    It is useful if the polynomial is in the form of a product linear expressions.
  6. Apr 16, 2013 #5


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    Do you suppose they were referring to the multiplicative inverse? ... such as getting the derivative of secant from cosine, or cotangent from tangent, etc.

    If y = sec(x), then
    ln(y) = -ln(cos(x))

    Differentiating gives:

    [itex]\displaystyle \frac{y'}{y}=-\frac{-\sin(x)}{\cos(x)}[/itex]


    It seems not much an improvement over just treating sec(x) as 1/cos(x) & using the chain rule .
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