Derivative of arcsec(x) and arccsc(x)

  • Thread starter NoOne0507
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  • #1
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I was trying to prove the derivatives of the inverse trig functions, but I ran into a problem when I tried doing it with arcsecant and arccosecant.

So the general process is this:

y = arcsec(x)
sec(y) = x
dy/dx * sec(y)tan(y) = 1
dy/dx = 1/[sex(y)tan(y)]

sec(y) = x
And for tan(y) we use the Pythagorean identities:

tan^2(y) = sec^2(y) - 1
tan(y) = [sec^(y) - 1] ^(1/2)

So dy/dx = 1/[x(x^2-1)^(1/2)]

However, my calculus book has one minor difference in it's derivative, an absolute value:
dy/dx = 1/[abs(x)(x^2-1)^(1/2)]

Where does this absolute value come from?
 

Answers and Replies

  • #2
AlephZero
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You missed the fact that [itex]\tan^2 y = \sec^2 y - 1[/itex] and therefore
[itex]\tan y = \pm \sqrt{\sec^2 y - 1}[/itex]

But only one of the "[itex]\pm[/itex]" solutions is the correct one, depending on which quadrant the angle is in.

The "abs(x)" gives the correct solution, if you always use the positive value of the square root.
 
  • #3
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A similar situation involving absolute value is
Integral of 1/x dx = Ln |x| + C

The absolute value is added because one can not take the log of a negative number.

The absolute value seems to come out of nowhere but it is just a way of using Math symbols to say what the words would say.
 

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