Substituting these identities into the formula for the derivative, we get 1 x x2 − 1− −−−−√ 1 x x 2 1 . However, the actual answer is the absolute value of that. I can't figure out where I am assuming x x is positive. It has been a while since I have mucked about with this kind of thing, so my apologies if this is a silly question.
y = arcsec x y = arcsec x can be defined in two ways. The first restricts the domain of sec y sec y to [0, π], y ≠π 2 [0, π], y ≠π 2. So the range of y y goes between [0, π2) ∪ (π2, π] [0, π 2) ∪ (π 2, π] and the slope of the function is always positive. The derivative is
You could try to derive the derivative of, say $f (x)=\mathrm {arcsec} (x^7)$ by finding the derivative of $\sec (f (x))$, applying the Chain Rule, and then solving for $f' (x)$, but in the long run it is probably going to be less work to remember the derivatives of the inverse trig functions than to derive them from scratch every time you need ...
The original question asked for the derivative of $\sqrt {\arcsec x}$; I managed to bring it till here after multiplying the denominator and numerator by conjugate of numerator and eliminating the square root.
0 Ted Shifrin has already explained the problem with your derivation, but if you're looking for a 'safer' way of finding the derivative then consider this: if y = arcsec x y = arcsec x, then using the identity arcsec x = arccos 1/x arcsec x = arccos 1 / x, we see that
The first method gives you the correct derivative for all v v where sec v sec v is defined. The second method, as written, is only applicable to 0 ≤ v ≤ π 0 ≤ v ≤ π (excluding v = π/2 v = π / 2), but in this interval, sec v tan v ≥ 0 sec v tan v ≥ 0 anyways, and so the absolute value can be dropped. However, in other intervals for v v, the formula v = arcsec(2u) v = arcsec (2 u ...
1 The arc secant function isn't defined in [−1, 1] [1, 1] so that the derivative must be studied separately for positive and negative x x. But we have arcsec(−x) = arcsec(x) + π arcsec (x) = arcsec (x) + π so that −arcsec′(−x) =arcsec′(x). arcsec (x) = arcsec (x)
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