### Video Transcript

If π¦ equals eight cot π₯ plus five
sec π₯, find ππ¦ by ππ₯ at π₯ equals π by six.

Weβre looking for the derivative of
ππ¦ by ππ₯. Well, actually weβre looking for
the value of ππ¦ by ππ₯ at π₯ equals π by six. But to find this, we first need to
find ππ¦ by ππ₯ in terms of π₯. So letβs differentiate. Weβre looking for the derivative of
eight cot π₯ plus five sec π₯. And we use the fact that the
derivative of a sum of functions is the sum of their derivatives and the fact that
the derivative of a number times a function is that number times the derivative of
the function to write this derivative ππ¦ by ππ₯ in terms of the derivatives of
cot π₯ and sec π₯.

The main part of this question is finding the derivative of these two trigonometric functions: cot and sec. There are
six trigonometric functions that we tend to use: sin, cos, tan, csc, sec, and cot. And certainly we could memorize all six of their derivatives if we wanted to, and
this isnβt a terrible idea. It would allow us to differentiate any trigonometric
function very quickly. Certainly if we memorized the
derivatives of sec and cot, then weβd pretty much be done with this question.

But if you have forgotten the
derivatives of sec and cot or if youβve never memorized them in the first place, all
is not lost. The key thing to do is to remember
the first two derivatives: the derivative of sin π₯ with respect to π₯ is cos π₯,
and the derivative of cos π₯ with respect to π₯ is minus sin π₯. From these two derivatives, we can
derive the other four. Letβs see how we can find the
derivative of cot when we know the derivatives of sin and cos.

Okay, so weβre looking for the
derivative of cot π₯ with respect to π₯. Now what is cot π₯? Well, you might know that itβs one
over tan π₯, and so weβre looking for the derivative of one over tan π₯ with respect
to π₯. And as tan π₯ is sin π₯ over cos
π₯, cot π₯ being the reciprocal of this must be cos π₯ over sin π₯. Now weβve succeeded in writing cot
π₯ in terms of sin and cos.

In fact, weβve shown that cot π₯ is
the quotient of cos π₯ and sin π₯. And so, we can differentiate it
using the quotient rule. So hereβs the quotient rule, and
now we can apply it setting π of π₯ to be cos π₯ and π of π₯ to be sin π₯. Now we have the derivative weβre
looking for written in terms of sin π₯ cos π₯ and their derivatives.

And we know the derivatives of sin
π₯ and cos π₯, so we can substitute them in. The derivative of cos π₯ is minus
sin π₯ and so the first term in the numerator becomes sin π₯ times minus sin π₯ or
minus sin π₯ squared. How about the second term in the
numerator? Well, the derivative of sin π₯ with
respect to π₯ is cos π₯. And so, the second term becomes cos
π₯ times cos π₯ or cos π₯ squared.

Thereβs not much we can do to the
denominator. It stays as sin π₯ squared, which
we can eventually write as sin squared π₯. But notice the numerator is minus
sin squared π₯ minus cos squared π₯, which is minus sin squared π₯ plus cos squared
π₯. And hopefully we recognize that we
can apply the identity sin squared π₯ plus cos squared π₯ equals one, which means
our numerator is negative one. And so the derivative of cot π₯
with respect to π₯ is negative one over sin squared π₯.

And taking the minus sign outside
the fraction, we get negative one over sin squared π₯. We can apply this result to the
derivative we wanted to find. The first term of ππ¦ by ππ₯ is,
therefore, eight times negative one over sin squared π₯. Before we move on, finding the
derivative of sec π₯ with respect to π₯ and hence the second term of ππ¦ by ππ₯,
letβs first recap how we found the derivative of cot π₯ with respect to π₯.

We first wrote cot π₯ in terms of
sin π₯ and cos π₯. And then seeing that we had a
quotient, we applied the quotient rule using the derivatives of sin π₯ and cos π₯
with respect to π₯, which we memorized. And simplifying by applying the
identity sin squared π₯ plus cos squared π₯ equals one gave us the simple expression
negative one over sin squared π₯. So now letβs follow the same steps
to find the derivative of sec π₯ with respect to π₯.

We can write sec π₯ in terms of sin
π₯ and cos π₯. In fact, we only need cos π₯. Sec π₯ is one over cos π₯. And we differentiate this by
applying the quotient rule with π of π₯ equal to one and π of π₯ equal to cos
π₯. The derivative of one with respect
to π₯ is just zero, and so the first term in the numerator vanishes. And so weβre left with just the
second term: minus one times the derivative of cos π₯ with respect to π₯. And we know what the derivative of
cos π₯ is with respect to π₯.

Itβs minus sin π₯. So combining this with the minus
one, we get sin π₯ in the numerator and of course cos squared π₯ in the
denominator. And now that we found the
derivative of sec π₯ with respect to π₯, we can apply this to our problem. The second term of ππ¦ by ππ₯
becomes five sin π₯ over cos squared π₯. Okay, now we can clear away our
working. Now weβve found ππ¦ by ππ₯ in
terms of π₯. We now just need to evaluate this
when π₯ is π by six.

Now we replace π₯ by π by six and
we get some expression involving sin π by six and cos π by six. And as π by six is a special
angle, hopefully we know the values of sin π by six and cos π by six. Sin π by six is a half and cos π
by six is root three over two. We can substitute these values and
simplify. For example, a half squared is a
fourth. And root three over two squared is
three-fourths. Negative one over a fourth is
negative four, and a half over three-fourths is two-thirds.

And so we get eight times negative
four plus five times two-thirds. Simplifying, we find that the value
of ππ¦ by ππ₯ at π₯ equals π by six is negative 86 over three. This problem required us to find
the derivatives of cot and sec. We found that we could work out
what these derivatives were using the known derivatives of sin and cos and also the
quotient rule. The derivative of cot π₯ with
respect to π₯ is [minus] one over sin squared π₯, which can also be written as minus csc
squared π₯. And the derivative of sec π₯ with
respect to π₯ we found to be sin π₯ over cos squared π₯, which can also be written
as sec π₯ times tan π₯.

However, we left these derivatives
in terms of sin and cos to make it easy to evaluate them at π₯ equals π by six. We donβt need to memorize the
derivatives of all six trigonometric functions, although it might be a good idea to
do so as long as we remember the derivatives of sine and cos. To be fair, even the derivatives of
sin and cos could be worked out from other more basic facts. But itβs definitely worth
memorizing the derivatives of sin and cos at the very least.