Welcome to a proof
of the derivative of f
of x equals arctangent x.
We'll prove the derivative of arctangent x
with respect to x equals one divided
by the quantity one plus x squared.
To begin, we'll let y equal arctangent x,
where remember x will be
the tangent function value,
and y would be the angle,
and then it follows that we
can rewrite this equation
as tangent y equals x where the angle y
would be on the open interval from
negative pi over two to
positive pi over two,
meaning the angle y would be in
the first quadrant or the fourth quadrant,
and let's model angle y using a triangle
where we can say tangent y is equal to x
if we want x over one,
and since tangent theta
is equal to the ratio of the opposite side
to the adjacent side of a right triangle,
using this right triangle,
if we label this angle y,
we can label the opposite side
x and the adjacent side one,
and therefore using the
Pythagorean Theorem,
we can determine the
length of the hypotenuse
would be the square root of the quantity
one squared plus x squared
or just the square root
of the quantity one plus x squared.
However, one important thing to recognize
when using this triangle
to model the angle y
given the tangent y equals
x, notice how if the angle y
was in the first quadrant,
then the tangent function value
would be positive so x would be positive,
but if the angle y is
in the fourth quadrant,
then the tangent function
value would be negative
and x would be negative.
Now using this triangle,
notice how we can also say
that sine y would be equal to the ratio
of the opposite side of the hypotenuse,
or x divided by the square root
of the quantity one plus x squared.
We can also say that cosine y
would be equal to the
ratio of the adjacent side
to the hypotenuse, which
would be one divided
by the square root of the
quantity one plus x squared.
Now for the next step in our proof,
we'll differentiate both
sides of this equation here
with respect to x, and
notice how because we have
an implicit equation,
we do have to perform
implicit differentiation, so now we have
the derivative of tangent y with respect
to x equals derivative
of x with respect to x
and the derivative of
tangent y with respect
to x would be equal to
secant squared y times dy/dx
and then we have equals the derivative
of x with respect to x
which is equal to one.
And now solving for
dy/dx, you would divide
both sides by secant
squared y, so we have dy/dx
equals one divided by secant squared y,
however, since one over secant y is equal
to cosine y, one over secant squared y
is equal to cosine squared y.
Remember we already know that cosine y
is equal to one divided by the square root
of the quantity one plus x squared
and therefore we have
dy/dx equals one divided
by the square root of the
quantity one plus x squared,
all squared, erase the
second power, which gives us
our derivative formula.
Dy/dx equals one divided by the quantity,
one plus x squared.
Notice how for any value x, we're going
to have a positive
derivative function value
which means if we take a look at the graph
of f of x equals arctangent x,
the slope of any tangent
line will be positive
and the function is always increasing.
Let's go ahead and verify
this before we finish.
So here's the graph of f
of x equals arctangent x.
Notice how at any x value, the slope
of the tangent line will be positive
and this function is increasing
over its entire domain, or we can say
it's monotonically increasing.
I hope you found this helpful.
