We're asked to determine
the derivative function,
then determine the equation of the tangent line
at the point pi, comma, negative four pi
times E to the power of pi.
Notice the function F of X
is a product of three functions of X
and therefore we'll have to apply the product rule,
within the product rule.
Let's let the first function be 4X, E to the X,
and that the second function be secant X.
So we'll let U equal 4X, E to the X
and V equal secant X
and now applying the product rule,
F prime of X
is equal to the first function
of 4X, E to the X,
times the derivative of the second function
which is a derivative of secant X
with respect to X.
And then plus the second function of secant X
times the derivative of the first function,
which is a derivative
of 4X, E to the X,
with respect to X.
Just notice how to find this derivative,
we will have to apply the product rule again.
For the next step,
we have F prime of X,
equals 4X, E to the X,
times the derivative of secant X,
which is equal to secant X, tangent X.
And then we have plus secant X,
times the derivative of 4X, E to the X.
And again, here we have to apply the product rule,
where the first function is going to be 4X,
and the second function is going to be E to the X.
So the derivative is equal to the first function of 4X,
times the derivative of E to the X, with respect to X,
which is E to the X,
plus the second function of E to the X,
times the derivative of the first function,
which is the derivative of 4X with respect to X,
which is four.
And now we need to simplify.
We have F prime of X equals 4X, E to the X,
secant X, tangent X,
plus distributing,
we have 4X, E to the X, secant X,
and then plus 4E to the X, secant X.
Let's go ahead and factor out
the greatest common factor.
Notice all three products have a factor of four,
a factor of E to the X
and a factor of secant X.
We can write F prime of X
as F prime of X equals 4E to the X, secant X,
and then times the quantity
X tangent X,
plus X,
plus one.
Notice if we distribute,
we still have the same expression.
So this is the first part of the question,
we found the derivative function.
And now from here,
we want to determine the equation of the tangent line
at the point pi, comma, negative four pi,
times E to the power of pi.
So we need to find the slope of the tangent line
by evaluating the derivative function
at X equals pi.
So F prime of pi
is equal to 4E to the pi,
times secant pi,
times a quantity pi,
times tangent pi,
plus pi, plus one.
Now need to find the value
of secant pi and tangent pi.
Secant pi is equal to negative one
and tangent pi is equal to zero.
Let's verify this on the unit circle.
On the unit circle, secant theta
is equal to one divided by X,
the terminal side of pi radians
intersects unit circle of the point
negative one zero.
The secant function value
is one divided by negative one or negative one,
and tangent theta is equal to Y divided by X,
which gives us zero divided by negative one,
which is zero.
This gives us four times E to the pi,
times negative one,
times a quantity pi times zero plus pi plus one.
Simplifying we have negative four,
E to the pi,
times the quantity pi plus one.
So this is a slope of the tangent line at the given point.
So now we can find the equation of the tangent line
using the slope and the point,
using point-slope form.
So let's go ahead and do that.
Remember, point-slope form is
Y minus Y one, equals M
times a quantity X minus X one,
where in our case X one comma Y one
is the given point,
and the slope is a derivative function value.
So we have Y minus the Y value of negative four pi,
times E to the power of pi,
equals the slope of negative four E to the pi,
times the quantity pi plus one,
times the quantity X minus X one,
which gives us X minus pi.
Leaving this in point-slope form,
we have the equation,
Y plus four pi, E to the pi,
equals negative 4E to the pi,
times a quantity pi plus one,
times a quantity X minus pi.
And if they want to solve for Y,
we would simply subtract four pi, E to the pi
on both sides of the equation.
And I'll go ahead and show that as well.
Solving for Y,
we would have Y equals negative four, E to the pi,
times a quantity pi plus one,
times a quantity X minus pi,
and then minus four pi times E to the power of pi.
Either form of the linear equation
is the equation of the tangent line.
And let's verify this graphically.
In blue, we have the graph of the given function,
the point pi, comma negative four pi,
times E to the power of pi is this red point.
And we just found the equation of this red line,
the tangent line,
which we can see is tangent to the curve
at the given point.
I hope you found this helpful.