Welcome to our proof of
the derivative of f of x equals arccotangent x.
We'll prove the derivative of arccotangent x,
with respect to x, equals negative,
one divided by the quantity, one plus x-squared.
So to begin, we'll let y equal arccotangent x,
and notice in this equation,
x would be the cotangent function value,
and y would be the angle.
And then, it follows that cotangent y must equal x,
where the angle y would be on the open integral
from zero to pi radians, meaning the angle, y,
can be in the first or second quadrant.
Let's go ahead and model angle, y,
using a reference triangle, or a triangle
where we have cotangent y equals x,
so if we want, x over one,
and since cotangent theta is equal to the ratio of the
adjacent side to the opposite side,
using this right triangle here, if we label this angle y,
then we can label the adjacent side, x,
and the opposite side, one, and therefore,
using the pythagorean theorem,
we can determine the length of the hypotenuse
would be equal to the square root of the quantity
one squared plus x-squared, which simplifies to
the square root of the quantity, one plus x-squared.
However, it is important to recognize that
if angle y is in the first quadrant,
then the cotangent function value would be positive,
so x would be positive,
but if the angle y is in the second quadrant,
then the cotangent function value would be negative,
and therefore, x would be negative.
Also notice, using this right triangle,
that sine y, which is equal to the ratio
of the opposite side to the hypotenuse,
would be equal to one, divided by,
the square root of the quantity, one plus x-squared,
and cosine y, which is equal to the ratio
of the adjacent side to the hypotenuse,
would be equal to x, divided by
the square root of the quantity, one plus x-squared.
So, for the next step in our proof,
we'll differentiate both sides of this equation here,
with respect to x, and because we have an implicit equation,
we do have to perform implicit differentiation.
So now, we have the derivative of cotangent y,
with respect to x, equals
the derivative of x, with respect to x.
The derivative of cotangent y, with respect to x,
is equal to negative cosecant squared y, times dy/dx,
which must equal the derivative of x,
with respect to x is one, and now, we solve for dy/dx
by dividing both sides by negative cosecant squared y,
so we have dy/dx equals negative one, divided by
cosecant squared y, and because one over cosecant y
is equal to sine y, we can rewrite this as
negative sine squared y.
And now, because we know that sine y is equal to
one, divided by the square root of the quantity,
one plus x-squared, we have dy/dx equals, notice how here,
the negative is not attached to the exponent,
so we have a negative outside the parentheses,
and then, we have one, divided by the square root
of one plus x-squared, all squared,
or raised to the second power, which equals negative one,
divided by the quantity, one plus x-squared.
And therefore, we have our proof.
Also notice how for any x value,
the derivative function value is going to be negative,
which means for the function, f of x equals arccotangent x,
at any x value, the slope of the tangent line
is going to be negative,
and the function is always decreasing.
Let's verify this by looking at the graph
of f of x equals arccotangent x.
Here's a graph of f of x, and again,
because the derivative function value is always negative,
notice how at any x value,
the slope of the tangent line is going to be negative,
and this function is always decreasing,
again, because the first derivative is always negative,
so they can say that this function is
monotonically decreasing.
I hope you found this helpful.