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Derivative of an inverse function

  1. Dec 3, 2016 #1
    1. The problem statement, all variables and given/known data
    $$\int\frac{dx}{x\sqrt{x^2-1}}=\left\{ \begin{array} {lc} \sec^{-1}(x)+C_1 & {\rm if}~x>1 \\ \sec^{-1}(-x)+C_2 & {\rm if}~x<-1 \end{array} \right.$$
    Why the second condition ##\sec^{-1}(-x)+C_2~~{\rm if}~x<-1## ?
    Snap1.jpg

    2. Relevant equations
    Derivative of inverse secant:
    $$(\sec^{-1}(x))'=\frac{1}{x(\pm\sqrt{x^2-1})}$$

    3. The attempt at a solution
    The derivative is positive from -∞ to +∞ (the green lines). as i understand i have to make the function ##\sec^{-1}(-x)## positive when x<-1, but if ##y=\sec^{-1}(x)## is defined like in the drawing it is positive alsways
     
  2. jcsd
  3. Dec 3, 2016 #2
  4. Dec 7, 2016 #3
    Continuity: ##\sec^{-1}(-x)=\pi-\sec^{-1}(x)## so:
    $$\int\frac{dx}{x\sqrt{x^2-1}}=\left\{ \begin{array} {lc} \sec^{-1}(x)+C_1 & {\rm if}~x>1 \\ -\sec^{-1}(x)+C_2 & {\rm if}~x<-1 \end{array} \right.$$
    But that gives a negative derivative, if i substitute x<-1 in:
    $$\frac{dx}{x\sqrt{x^2-1}}$$
    Snap1.jpg
     
  5. Dec 8, 2016 #4

    vela

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    If ##x<-1##, then
    $$\frac{dx}{x\sqrt{x^2-1}}= \frac{d(-x)}{(-x)\sqrt{(-x)^2-1}} = \frac{d(-x)}{\lvert (-x)\rvert \sqrt{(-x)^2-1}} = d[\sec(-x)]$$
     
  6. Dec 8, 2016 #5
    I don't understand the transition:
    $$\frac{d(-x)}{(-x)\sqrt{(-x)^2-1}} = \frac{d(-x)}{\lvert (-x)\rvert \sqrt{(-x)^2-1}}$$
    Why did (-x) in the denominator became |(-x)|? why is it allowed?
     
  7. Dec 8, 2016 #6

    vela

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    ##x## is negative, no?
     
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