### Video Transcript

If π¦ is equal to the square root
of negative four π§ plus nine and π§ is equal to the sec of six π₯, find the dπ¦ by
dπ₯ at π₯ is equal to π by 18.

The question wants us to find dπ¦
by dπ₯ which is the first derivative of π¦ with respect to π₯ when π₯ is equal to π
by 18. To do this, we need to first find
an expression for dπ¦ by dπ₯ and then substitute π₯ is equal to π by 18 into this
expression. Normally, we would just do this by
differentiating our expression for π¦ with respect to π₯. However, we can see a problem. Weβre not given π¦ as a function in
π₯. Instead, weβre given π¦ as a
function in π§.

And we can then see π§ in turn is a
function in π₯, so we can do this by using the chain rule. So, letβs start by recalling the
chain rule. The chain rule tells us if π¦ is
some function of π§ and π§ in turn is some function in π₯, then dπ¦ by dπ₯ will be
equal to dπ¦ by dπ§ multiplied by dπ§ by dπ₯. And this is exactly what we have
here. So, to find our expression for dπ¦
by dπ₯, we need to differentiate π¦ with respect to π§ and π§ with respect to
π₯.

Letβs start by finding an
expression for dπ¦ by dπ§. Thatβs the derivative of the square
root of negative four π§ plus nine with respect to π§. Thereβs a few different ways of
differentiating this. For example, this is the
composition of two functions. So, we could differentiate this by
using the chain rule. However, weβre going to use the
general power rule.

To do this, we first need to recall
that taking the square root is the same as raising a number to the power for
one-half. So, we can rewrite this as negative
four π§ plus nine all raised to the power of one-half. Then, we can evaluate this
derivative by using the general power rule. We recall this tells us for the
differentiable function π of π§ in any real constant π, the derivative of π of π§
all raised to the πth power with respect to π§ is equal to π times π prime of π§
multiplied by π of π§ all raised to the power of π minus one.

We can do this directly. First, we need to multiply by our
exponent π. In this case, we can see this is
one-half. Next, in the power rule for
differentiation, we need to differentiate our inner function. In this case, our inner function is
a linear function. We know the derivative of this will
be the coefficient of π§, which is negative four. Finally, in the general power rule,
we need to multiply it by our original function. However, we need to reduce the
exponent by one. So, we need to multiply it by
negative four π§ plus nine all raised to the power of one-half minus one.

And we can simplify this
expression. First, in our coefficient, we have
one-half multiplied by negative four, which we can calculate as negative two. Next, in our exponent, we have
one-half minus one, which is, of course, just equal to negative one-half. So, we can simplify our expression
for dπ¦ by dπ§ to be negative two times negative four π§ plus nine all raised to the
power of negative one-half.

In fact, we can simplify this
further. By using our laws of exponents,
instead of multiplying by a number raised to the power of negative one-half, we can
divide by the square root of this number. So, we can rewrite this as negative
two divided by the square root of negative four π§ plus nine.

And thereβs one more thing weβre
going to do. Remember, weβre going to use the
chain rule to find an expression for dπ¦ by dπ₯. After that, weβre going to
substitute π₯ is equal to π by eight π§ into this expression. And to make this easier, we would
want our expression for dπ¦ by dπ§ to be in terms of π₯. To do this, weβre going to use our
substitution π§ is equal to the sec of six π₯. And by doing this, we get dπ¦ by
dπ§ is equal to negative two divided by the square root of negative four times the
sec of six π₯ plus nine.

Now that we found an expression for
dπ¦ by dπ§, we need to find an expression for dπ§ by dπ₯. Thatβs the derivative of the sec of
six π₯ with respect to π₯. And to do this, we need to recall
one of our standard rules for differentiating reciprocal trigonometric
functions. For any real constant π, the
derivative of the sec ππ₯ with respect to π₯ is equal to π times the tan of ππ₯
multiplied by the sec of ππ₯.

In this case, we can see our value
of the constant π is six. So, by setting π equal to six, we
get dπ§ by dπ₯ is equal to six tan of six π₯ multiplied by the sec of six π₯. And now that we found the
expressions for dπ¦ by dπ§ and dπ§ by dπ₯, we can substitute these into our chain
rule to find dπ¦ by dπ₯. So, we multiply our expressions for
dπ¦ by dπ§ and dπ§ by dπ₯ together. We get dπ¦ by dπ₯ is equal to
negative two multiplied by six tan of six π₯ times the sec of six π₯ all divided by
the square root of negative four times the sec of six π₯ plus nine.

And We can simplify this. In our numerator, we have two times
six, which is equal to 12. But remember, the question is not
just asking us to find an expression for dπ¦ by dπ₯. We need to find dπ¦ by dπ₯ at π₯ is
equal to π by 18. So, we need to substitute π₯ is
equal to π by 18 into this expression. And we could do this directly. However, we can simplify this. We notice all of our arguments are
six π₯. And we can calculate six times π
by 18 is equal to π by three. So, when we substitute π₯ is equal
to π by 18 into this expression, all of our arguments will be π by three.

Substituting π₯ is equal to π by
18 into this expression and simplifying, we get negative 12 times the tan of π by
three multiplied by the sec of π by three all divided by the square root of
negative four times the sec of π by three plus nine. And now, we can just evaluate this
expression. First, the tan of π by three is
equal to the square root of three. Next, we can see the sec of π by
three appears twice in this expression. And we know the sec of π by three
will be equal to one divided by the cos of π by three. But the cos of π by three is a
half. And one divided by one-half is just
equal to two.

So, we can replace both instances
of the sec of π by three with two. This gives us negative 12 times
root three times two all divided by the square root of negative four times two plus
nine. And we can just start calculating
this expression. First, in our denominator, negative
four times two is negative eight, plus nine is equal to one. So, our denominator is the square
root of one, which is just equal to one. And in our numerator, we have 12
multiplied by two, which is equal to 24. So, we can just calculate this
expression to give us negative 24 root three, which is our final answer.

Therefore, we were able to show if
π¦ is equal to the square root of negative four π§ plus nine and π§ is equal to the
sec of six π₯, then dπ¦ by dπ₯, when π₯ is equal to π by 18, is equal to negative
24 root three.