Welcome to a proof
of the derivative of f
of x equals arcsine x.
We'll prove the derivative of arcsine x
with respect to x equals one divided
by the square root of the
quantity one minus x squared.
We begin our proof, let y equal arcsine x,
so notice in this equation,
x would be the sine function
value and y would be the angle,
and therefore it follows
that we can rewrite
this equation as sine y
equals x where y the angle
would be on the closed
interval from negative pi
over two to positive pi over two.
For our next step, let's model angle y
using a right triangle where
if we have sine y equals x
so if we want x over one,
since sine theta is equal to the ratio
of the opposite side to the hypotenuse
of any right triangle, we can label
the opposite side x and the hypotenus one.
So again, this is our angle y
and because sine y is equal to x,
we can label the opposite side x
and the hypotenuse one and therefore using
the Pythagorean Theorem, we can label
the adjacent leg, the square root
of the quantity one minus x squared.
So using this triangle,
we know that sine y
is equal to x and we
also know that cosine y
which is equal to the
ratio of the adjacent side
to the hypotenuse,
would be the square root
of the quantity one minus x squared.
However, it should be noted
that the sine function value
can be positive or
negative when the angle y
is on the closed 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.
If the angle's in the first quadrant,
then the sine function
value would be positive,
and if the angle is in
the fourth quadrant,
the sine function value would be negative.
However, the adjacent side of the triangle
would always be positive in the first
and fourth quadrants
because the adjacent side
would be the x coordinate
on the coordinate plane
which is positive in the
first and fourth quadrants.
So for the next step in our proof,
we use implicit differentiation
and differentiate both
sides of the equation here
with respect to x.
So now we have the derivative of sine y
with respect to x equals the derivative
of x with respect to x.
The derivative of sine y with respect to x
would be cosine y times dy/dx, you notice
we have an extra factor of
dy/dx because of the chain rule,
and then on the right side
we have the derivative
of x with respect to x is equal to one
and now solving for dy/dx
we'd divide both sides
by cosine y, so we have
dy/dx equals one divided
by cosine y, and since we know cosine y
is equal to the square
root of the quantity
one minus x squared, we
have dy/dx equals one over
the square root of the
quantity one minus x squared.
We can also write one
over cosine y as secant y
using our reciprocal
identity and then referring
back to the triangle,
for secant y we'd have
the ratio of the hypotenuse
to the adjacent side,
which also gives us one
divided by the square root
of the quantity one minus x squared.
And therefore we have our proof,
the derivative of arcsine
of x with respect to x
equals one divided by the square root
of the quantity one minus x squared.
Notice wherever the derivative is defined,
this value would be positive which means,
wherever the derivative
function value is defined,
the slope of the tangent
line would be positive.
And let's verify this
by looking at the graph
of f of x equals arcsine of x.
So here's the graph of
f of x equals arcsine x.
Notice how the derivative
function value is going
to be defined except at
one and negative one,
but for every value in the
domain of the derivative
function, the derivative
function value would be positive
and therefore the slope
of the tangent line
to the function is positive
and the function is increasing.
I hope you found this helpful.
