If F of X equals two arctangent three X,
we want to find F prime of
X, the derivative function,
then find F prime of one,
which will give us the slope
of the tangent line at X equals one.
Notice how our function F of
X is a composite function,
where the inner function would be three X,
so we do have to apply the chain rule
when finding our derivative.
Looking at our derivative
from this here below,
the derivative formula on the left is
for the derivative of the basic function,
arctangent X, with respect to X,
but the second derivative
formula here on the right
also includes the chain rule,
where we have the
derivative of arctangent U,
with respect to X,
and therefore, U would
be the inner function.
Looking at our function,
notice how U would be equal to three X,
and we also have to
find U prime, or DU DX.
If U is equal to three X, then
U prime is equal to three.
Now, we can think of this
as finding the derivative
of two times arctangent U, and therefore,
F prime of X would be equal to two,
again, times the
derivative of arctangent U,
which would be equal to one divided by
the quantity one plus U squared,
where U is equal to three X,
so we have three X squared, times U prime,
where U prime is three.
Therefore, the derivative function
F prime of X would be equal to,
Notice how the numerator would
be two times three, or six,
and then we have divided
by the quantity one plus,
let's write three X to the
second as nine X squared.
This is the first part of the question.
This is our derivative function.
Now, for the second part, we want to
evaluate the derivative
function at X equals one.
F prime of one would be
equal to six divided by
the quantity one plus
nine times one squared.
Well, that would be six tenths,
which does simplify to three fifths,
since we have a common factor of two.
F prime of one equals three fifths,
which again, is a slope of the
tangent line at X equals one,
which we will now verify graphically.
The given function F of X
is graphed here in blue,
the point of the function
when X equals one
would be this green point here,
and this red line is the
tangent line at that point.
Since F prime of one is
equal to three fifths,
the slope of this tangent
line is three fifths,
which we can verify.
If we start at the Y intercept here,
and then we go up three units,
and right five units,
notice how we find another
point on our tangent line,
verifying the slope is three fifths,
and therefore, our derivative
function value is correct.
I hope you found this helpful.
