I Derivative of Inverse Secant

1. Oct 16, 2016

Please refer to the below image (Example 5).

Do anyone know how 5x^4 > 1 > 0?

Last edited: Oct 16, 2016
2. Oct 16, 2016

Math_QED

If $5x^4 < 1$ there would be a negative number under the root thus this is not part of the domain of the function.Therefore they assume $5x^4 \geq 1$.

3. Oct 16, 2016

I still don't understand. Could someone explain this please. So, what happen if there is a negative number under the root?

Last edited: Oct 16, 2016
4. Oct 16, 2016

Staff: Mentor

I suspect that the domain here are the real numbers, although not explicitly mentioned. But for $|u| < 1$ the root $\sqrt{u^2-1}$ becomes negative, and square roots of negative numbers like $\sqrt{-1}$ aren't real. $|u|=1$ is forbidden since the denominator would be zero.
For $u=5x^4$ this translates to the requirement $|u|=|5x^4|=5x^4 > 1$.

5. Oct 16, 2016

So, if $|u| < 1$ and $|u| = 1$ are forbidden, what the value of $|u|$ should be?

Is it $|u| > 1$?

If yes, isn't $|u| > 1$ is equal to $-1 > u > 1$ (from what I learned about inequality property)?

I still don't understand.

6. Oct 16, 2016

Staff: Mentor

Yes.
No.

$-1 > u > 1$ doesn't make sense. Smaller than $-1$ cannot be greater than $1$.
$|u| > 1$ means: $u > 1$ if $u$ is positive and $-u>1$ if $u$ is negative, have a look
https://en.wikipedia.org/wiki/Absolute_value

7. Oct 18, 2016

zinq

The explanation lies in the definition of sec-1(x) and the reason for this definition.

First we need to understand the graph of y = sec(x). Check out for example the 5th graph down on this page: http://cs.bluecc.edu/calculus/trigtools/graphs/trigGraphs/index.html [Broken].

In order to define an inverse function, we first need to find a portion of the domain of sec(x) on which sec(x) does not take the same value more than once. Ideally the function will also take all possible values at least once on this selected portion of the domain. This fussing is important!

A convenient portion of the domain of sec(x) for this purpose is the union of the two half-open intervals [0,π/2) and (π/2, π] — in other words the set X given by

X = [0, π/2) ∪ (π/2, π]​

— and this is what the standard convention is for obtaining the inverse function. Note that since sec(x) = 1/cos(x) by definition, and cos(x) takes values only between -1 and +1, it follows that sec(x) takes values only in the range

Y = (-∞, -1] ∪ [1, ∞).​

Now, to get the graph of the inverse function we need to consider the graph of y = sec(x) only on the restricted domain X, and then interchange the x- and y-axes. (This has the effect of flipping the graph about the 45° line y = x.) Thus the inverse function sec-1(x) has as its domain the set Y above, and takes values in the set X above.

Thus for all x where sec-1(x) makes sense, we have either x ≤ -1 or x ≥ +1. So x2 ≥ 1 in either case, which shows why the expression inside of the square root sign in the original problem is always greater than 0.

Last edited by a moderator: May 8, 2017
8. Nov 3, 2016

How do I know if x > 1 or x < -1?

9. Nov 3, 2016

Staff: Mentor

I know you don't mean what you said. If |u| < 1, then $u^2 - 1 < 0$, so $\sqrt{u^2 - 1}$ isn't real.

10. Nov 3, 2016

Staff: Mentor

By which branch of the graph of y = sec-1(x) you're on. See the graph here: http://www.wolframalpha.com/input/?i=y=arcsec(x)

In the first plot, change the plot type from Complex-valued plot to Real-valued plot.

BTW, arcsec(x) is just different notation for sec-1(x).

11. Nov 4, 2016

Lance219

$5x^4=sec y = \frac 1 {cos y}$, because |cos y| <=1, we have $|sec y| = |5x^4| = 5x^4 >= 1.$