### Video Transcript

Find the first derivative of π¦
where π¦ is equal to 16 csc of three π₯ multiplied by the sec of four π₯.

The question wants us to find the
first derivative of π¦, and it gives us an expression for π¦. We can see that π¦ is a function in
π₯. So when the question asks for the
first derivative of π¦, it means the first derivative of π¦ with respect to π₯. And thereβs actually a lot of
different ways of evaluating this expression. For example, we could use our
trigonometric identities to rewrite the csc of three π₯ and the sec of four π₯ in
our denominator as the sin of three π₯ and the cos of four π₯, respectively. We could then differentiate this by
using the product rule and the chain rule or the product rule and the general power
rule or the product rule and the quotient rule. All of these methods would
work. However, thereβs actually an easier
method. We need to notice that we know how
to different the csc of three π₯ and the sec of four π₯.

This means that π¦ is written as
the product of functions we know how to differentiate. So we can directly differentiate
this by using the product rule. So letβs start by recalling the
product rule. If π¦ is the product of two
differentiable functions, π of π₯ multiplied by π of π₯, then π¦ prime, the
derivative of π¦ with respect to π₯, will be equal to π prime of π₯ times π of π₯
plus π prime of π₯ times π of π₯. And we can see that our expression
for π¦ is in exactly this form. We set π of π₯ to be 16 times the
csc of three π₯ and π of π₯ to be the sec of four π₯.

Now, to use the product rule, we
just need to find expressions for π prime of π₯ and π prime of π₯. Letβs start with π prime of
π₯. Thatβs the derivative of 16 csc of
three π₯ with respect to π₯. The easiest way to evaluate this
derivative is to recall the following result for differentiating reciprocal
trigonometric functions. For any real constant π, the
derivative of the csc of ππ₯ with respect to π₯ is equal to negative π times the
csc of ππ₯ multiplied by the cot of ππ₯. In this case, we can see the value
of our constant π will be three. So by setting our value of π equal
to three and remembering we need to multiply by 16, we get π prime of π₯ is equal
to 16 multiplied by negative three csc of three π₯ multiplied by the cot of three
π₯.

And of course, we can simplify
this. 16 multiplied by negative three is
equal to negative 48. So we were able to find an
expression for π prime of π₯. Letβs now move on to finding an
expression for π prime of π₯. Thatβs the derivative of the sec of
four π₯ with respect to π₯. And we can evaluate this derivative
by recalling the following trigonometric derivative result. For any real constant π, the
derivative of the sec of ππ₯ with respect to π₯ is equal to π times the sec of
ππ₯ multiplied by the tan of ππ₯. This time, we can see our value of
the constant π is equal to four. So by setting the value of π equal
to four, we get π prime of π₯ is equal to four times the sec of four π₯ multiplied
by the tan of four π₯.

Now that we found expressions for
π prime of π₯ and π prime of π₯ and we know π of π₯ and π of π₯, weβre ready to
find an expression for π¦ prime. So weβll clear some space and
substitute these expressions into our product rule. This gives us the first derivative
of π¦ with respect to π₯ is equal to negative 48 csc of three π₯ times the cot of
three π₯ multiplied by the sec of four π₯ plus four sec of four π₯ times the tan of
four π₯ multiplied by 16 times the csc of three π₯. And we want to simplify this
expression. First, weβll take out the shared
factor of 16 from both terms and the shared factor of the csc of three π₯ from both
terms and also the shared factor of the sec of four π₯ from both terms.

So weβre taking out our factor of
16 times the csc of three π₯ multiplied by the sec of four π₯. Remember that weβve only taken out
a factor of 16 from our original coefficient, and we know negative 48 divided by 16
is negative three. So this gives us negative three
times the cot of three π₯. In our second term, weβre just left
with four times the tan of four π₯. And this just leaves us with the
following expression, and we could just leave our answer like this. However, thereβs one more piece of
simplification weβre going to do. Inside of our parentheses, weβre
going to rearrange our two terms. And this gives us our final
answer.

Therefore, we were able to show if
π¦ is equal to 16 csc of three π₯ times the sec of four π₯, then the first
derivative of π¦ with respect to π₯ is equal to 16 csc of three π₯ times the sec of
four π₯ multiplied by four tan of four π₯ minus three cot of three π₯.