### Video Transcript

If π¦ is equal to nine sin squared of π₯ all over five plus five cos of π₯, find dπ¦ by dπ₯.

Weβre asked to find dπ¦ by dπ₯. Thatβs the derivative of π¦ with respect to π₯. And we can see that π¦ is given as the quotient of two functions. And we know how to differentiate both of these functions, so weβll do this by using the quotient rule. We recall the quotient rule tells us, if π’ of π₯ and π£ of π₯ are differentiable functions and the derivative of π’ of π₯ over π£ of π₯ with respect to 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.

So to apply the quotient rule, we need to set π’ of π₯ to be the function in our numerator, thatβs nine sin squared π₯, and π£ of π₯ to be the function in our denominator, thatβs five plus five cos of π₯. And we see to use the quotient rule, weβre going to 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 nine sin squared of π₯ with respect to π₯.

And thereβs a few different ways of differentiating the sin squared of π₯. We could write it as the sin of π₯ multiplied by the sin of π₯ and use the product rule, we could use the chain rule, or we could use the double-angle formula for cosine. Any of these methods would work. However, weβre going to do this by using the general power rule.

We recall the general power rule tells us for differentiable function π of π₯ and 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. In this case, our inner function π of π₯ is the sin of π₯, and our exponent π is equal to two. Itβs also worth pointing out nine is a constant coefficient, so we can take this outside of our derivative. So, we have π is two and π of π₯ is the sin of π₯. However, to use the general power rule, we still need to find an expression for π prime of π₯. Thatβs the derivative of the sin of π₯ with respect to π₯, which we know is equal to the cos of π₯.

Now, we can use the general power rule to find an expression for π’ prime of π₯. We just need to substitute π is equal to two, π of π₯ is the sin of π₯, and π of π₯ is the cos of π₯ into our general power rule. We then need to multiply all of this by our coefficient of nine. This gives us π’ prime of π₯ is equal to nine times two cos of π₯ multiplied by the sin of π₯ raised to the power of two minus one. And this simplifies to give us 18 sin π₯ cos π₯.

However, weβre still not done. We still need to find an expression for π£ prime of π₯. Thatβs the derivative of five plus five cos of π₯ with respect to π₯. And we can just evaluate this derivative term by term. First, five is a constant. So, its rate of change with respect to π₯ will be equal to zero. Next, we know the derivative of five cos of π₯ with respect to π₯ will be equal to negative five times the sin of 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 18 sin of π₯ times the cos of π₯ multiplied by five plus five cos of π₯ minus negative five sin of π₯ times nine sin squared of π₯ all divided by five plus five cos of π₯ all squared. And thereβs a few different things we could do to simplify this. First, weβll take out the shared factor of nine sin of π₯ in our numerator. By doing this, we were able to rewrite dπ¦ by dπ₯ as the following expression.

And thereβs a few different things we can do to simplify this expression. First, in our denominator, we want to take out the factor of five from our square. But remember, because this is squared, this means we will end up having a factor of five squared. This means we can rewrite our denominator as five squared multiplied by one plus the cos of π₯ all squared. Next, we want to distribute two cos of π₯ over the inner parentheses in our numerator. And by doing this and simplifying, we get 10 cos of π₯ plus 10 times the cos squared of π₯. So, this gives us the following expression for dπ¦ by dπ₯. And it can be hard to see how weβre going to simplify this any further.

To simplify this, weβre going to need to notice the cos squared of π₯ plus the sin squared of π₯ is equal to one. We know this because of the Pythagorean identity. It might be easier to see how weβre going to use this by rewriting 10 cos squared of π₯ as five cos squared of π₯ plus five cos squared of π₯. Then, by using our Pythagorean identity, five cos squared of π₯ plus five sin squared of π₯ will just be equal to five. So, by using the Pythagorean identity, we were able to rewrite our expression for dπ¦ by dπ₯ as nine sin of π₯ times 10 cos of π₯ plus five times the cos squared of π₯ plus five all divided by five squared times one plus the cos of π₯ all squared.

The next thing we want to do is cancel the shared factor of five in our numerator and our denominator. This gives us the following expression. And at this point, it can be very difficult to see how weβre going to simplify this any further. We need to notice something interesting about two cos of π₯ plus the cos squared of π₯ plus one. We need to notice this is equal to one plus the cos of π₯ all squared. We can notice this by distributing the square over our parentheses and rearranging the terms. And in fact, one plus the cos of π₯ all squared is exactly what we have in our denominator, so we can cancel this shared factor out. So, by canceling this shared factor out, all weβre left with is nine over five times the sin of π₯. And this is our final answer.

Therefore, by using the quotient rule, we were able to show if π¦ is equal to nine sin squared of π₯ all over five plus five cos of π₯, then dπ¦ by dπ₯ would be equal to nine over five times the sin of π₯.