Welcome to a proof
of the derivative of f
of x equals arcsecant x.
We'll prove the derivative of arcsecant x
with respect to x
equals one divided by
the absolute value of x
times the square root of the
quantity x squared minus one.
To begin, we'll let y equal arcsecant x.
So notice in this equation,
x would be the secant function value
and y would be the angle.
And therefore it follows that
we can write the equation
secant y equals x where the angle y
would be on the closed
interval from zero to Pi
and the angle could not equal Pi over two
because secant is
undefined at Pi over two.
If we write this equation as
secant y equals x over one,
since secant theta is equal
to the ratio of the hypotenuse
to the adjacent side of a right triangle,
we can model angle y using
a reference triangle.
Though we do have to be careful though
because notice how when the angle y
is in the first quadrant,
x would be positive,
but when the angle y is
in the second quadrant,
x would be negative.
So we need to be a little careful
when we set up our triangle.
So if this is our angle y,
because the ratio of the
hypotenuse to the adjacent side
would be equal to x,
we might label the hypotenuse
x on the adjacent side one,
but because the hypotenuse
of a reference triangle
is always positive,
we're going to label the
hypotenuse the absolute value of x
and therefore if x is positive,
the adjacent side here
would be positive one
and if x is negative, we
would use negative one
on the adjacent side.
Then using the Pythagorean Theorem,
we can determine that the
opposite side would be
the principal square root of the quantity
x squared minus one.
We know the opposite side
of this reference triangle,
because the y-coordinate is positive
in the first and second quadrants.
Using this triangle, since sine theta
is equal to the ratio of the opposite side
of the hypotenuse,
we can write sine y is
equal to the square root
of the quantity x squared minus one
divided by the opposite value of x.
And for cosine y, instead of using
plus or minus one divided
by the absolute value of x,
because we know that
secant y is equal to x,
we know cosine y would
be the reciprocal of x
which would be one over x.
Again, here x can be positive or negative
based upon the quadrant.
And now for the next step in our proof,
we'll differentiate both
sides of this equation here
with respect to x.
Which because we have
an implicit equation,
requires implicit differentiation.
So the derivative of
secant y with respect to x
is equal to secant y times
tangent y times dy dx.
And this is equal to the
derivative of x with respect to x
which is equal to one.
And now solving for dy dx,
we would divide both sides
by secant y tangent y.
So we know dy dx is
equal to one divided by
secant y times tangent y.
And because one over secant
y is equal to cosine y,
and one over tangent y
is equal to cotangent y,
we can write this as
dy dx equals cosine y times cotangent y.
And because cotangent
y is equal to cosine y
divided by sine y,
we can write this product as
cosine y over one times cosine y over one
times one over sine y.
And now performing substitution.
We know cosine y is equal to one over x.
So we have one over x times one over x
and then times the
reciprocal of the sine y.
Well if this is sine y,
the reciprocal would be
the absolute value of x
divided by the square root of the quantity
x squared minus one.
Now before I multiply, we can simplify.
There's a common factor of x
between the numerator and the denominator
and we know the absolute value of x
is always going to be positive.
And x times x which equals x squared
will also always be positive.
So when we simplify out
the common factor of x,
to ensure the result is always positive,
we write the remaining factor of x
as the absolute value of x.
And we have our proof.
We have the derivative of arc secant x
with respect to x equals,
one divided by the absolute value of x
times the square root of the quantity
x squared minus one.
Notice how this derivative function value
is always going to be positive
regardless of whether x
is positive or negative.
Which means the slope of the tangent line
is always going to be positive.
F of x equals arc secant
x is always increasing
or monotonically increasing.
And let's verify this graphically.
So here's a graph of
y equals arc secant x.
Notice how anywhere where the
function is differentiable,
the slope of the tangent
line would be positive.
And the function is increasing
over its entire domain.
I hope you found this helpful.
