### Video Transcript

If π¦ is equal to the sin of three π₯ divided by negative three plus the cos of three π₯, find dπ¦ by dπ₯.

In this question, weβre asked to find the first derivative of π¦ with respect to π₯. And we can see that π¦ is the quotient of two trigonometric functions. Since trigonometric functions are differentiable across their entire domain, weβll do this by using the quotient rule. We recall the quotient rule tells us the derivative of the quotient of two functions π of π₯ over π of π₯ is equal to π prime of π₯ times π of π₯ minus π prime of π₯ times π of π₯ all divided by π of π₯ all squared. And this is valid at all values of π₯ where π is differentiable at π₯, π is differentiable at π₯, and π of π₯ is not equal to zero.

So to apply the quotient rule, weβll start by setting π of π₯ to be the function in our numerator β thatβs the sin of three π₯ β and π of π₯ to be the function in the denominator. Thatβs negative three plus the cos of three π₯. Then, weβve rewritten π¦ to be equal to π of π₯ divided by π of π₯. So we can differentiate this by using the quotient rule.

To apply the quotient rule, we need to find expressions for π prime of π₯ and π prime of π₯. Letβs start with π prime of π₯. To differentiate π of π₯, we need to differentiate the sin of three π₯. And we can differentiate this by recalling the following result. For any real constant π, the derivative of the sin of ππ₯ with respect to π₯ is equal to π multiplied by the cos of ππ₯. In our case, the value of π is three. So π prime of π₯ is equal to three times the cos of three π₯.

Now, we want to find the derivative of π of π₯. We can see that π of π₯ contains the cos of three π₯. And we can differentiate this by recalling the following result. For any real constant π, the derivative of the cos of ππ₯ with respect to π₯ is negative π times the sin of ππ₯. Weβll differentiate π of π₯ term by term. Weβll start with the first term negative three. Negative three is a constant; it doesnβt vary as π₯ varies. So its rate of change is going to be zero. This means we just need to differentiate the cos of three π₯. We do this by using our result with π equal to three. π prime of π₯ is equal to negative three times the sin of three π₯.

Weβre now ready to find an expression for dπ¦ by dπ₯ by using the quotient rule. Substituting in our expressions for π of π₯, π of π₯, π prime of π₯, and π prime of π₯, we get dπ¦ by dπ₯ is equal to three times the cos of three π₯ multiplied by negative three plus the cos of three π₯ minus negative three sin of three π₯ times the sin of three π₯ all divided by negative three plus the cos of three π₯ all squared.

We now want to simplify this expression. Letβs start with the first term in our numerator. Weβll distribute three cos of three π₯ over our parentheses to get negative nine cos of three π₯ plus three cos squared of three π₯. Next, weβll simplify the second term in our numerator. Negative one times negative three sin π₯ multiplied by sin of three π₯ is equal to three sin squared of three π₯. This gives us the following expression for dπ¦ by dπ₯.

And we could stop here. However, we can actually simplify this expression even further. And to do this, weβre going to need to recall the Pythagorean identity. We recall this tells us for any value of π, the sin squared of π plus the cos squared of π is equal to one. This is very similar to the last two terms in our numerator. First, weβll take out a factor of three from both of these terms. This gives us the following expression. Then, in the Pythagorean identity, weβll replace π with three π₯, giving us that the sin squared of three π₯ plus the cos squared of three π₯ is equal to one. Therefore, by using the Pythagorean identity, we can replace this with one. Doing this and then reordering the two terms in our numerator gives us our final answer.

If π¦ is equal to the sin of three π₯ divided by negative three plus the cos of three π₯, then dπ¦ by dπ₯ is equal to three minus nine cos of three π₯ all divided by negative three plus the cos of three π₯ all squared.