### Video Transcript

If π¦ is equal to one plus two times the sin squared of two π₯ all divided by three minus three sin squared of two π₯, find dπ¦ by dπ₯.

Weβre given π¦ as the quotient of two trigonometric functions and asked to find dπ¦ by dπ₯. Thatβs the derivative of π¦ with respect to π₯. So weβll do this by using the quotient rule. And we could do this directly right now. However, itβs always worth checking if we can simplify our expression. And in this case, we can. Weβll take out the common factor of three in our denominator. This would give us a new denominator of three times one minus the sin squared of two π₯. And we can simplify one minus the sin squared of two π₯ by using the Pythagorean identity.

Recall, the Pythagorean identity tells us the cos squared of π plus the sin squared of π is equivalent to one. Weβll subtract the sin squared of π from both sides of this equivalence. This gives us the cos squared of π is equivalent to one minus the sin squared of π. And we can make this exactly what we have in our denominator. We just need to set π equal to two π₯. So instead of having one minus the sin squared of two π₯ in our denominator, we can instead have the cos squared of two π₯. So by using the Pythagorean identity, we can rewrite π¦ as one plus two sin squared of two π₯ all divided by three times the cos squared of two π₯.

And now we could differentiate this by using the power rule for differentiation. However, we can actually make this even easier. Weβre going to divide both terms in our numerator separately by our denominator. Doing this gives us the following expression. And we can simplify both of these by using trigonometric identities. First, weβre dividing by the cos squared in our first term. And we need to remember dividing by the cos of π is the same as the sec of π. So we can rewrite this as one-third times the sec squared of two π₯.

We can do something very similar in our second term. Remember, sine divided by cosine is the tangent. So the sin squared of two π₯ divided by the cos squared of two π₯ will be the tan squared of two π₯. So we can rewrite the second term as two-thirds times the tan squared of two π₯. And now we could differentiate this term by term by using either the product rule, the chain rule, or the general power rule. And any of these methods would work and give us the correct answer. However, we can actually simplify this even further. We need to go back to our Pythagorean identity.

We want to use the Pythagorean identity to rewrite the tan squared of two π₯. So weβre going to divide our Pythagorean identity through by the cos squared of π. Dividing the Pythagorean identity through by the cos squared of π, we get the cos squared of π divided by the cos squared of π plus the sin squared of π over the cos squared of π is equivalent to one over the cos squared of π.

And we can simplify each of these terms. First, the cos squared of π over the cos squared of π is equal to one. Next, remember, the sin of π divided by the cos of π is the tan of π. So the sin squared of π divided by the cos squared of π is the tan squared of π. And finally, we need to remember the sec squared of π is equivalent to dividing through by the cos squared of π. So this gives us one plus the tan squared of π is equivalent to the sec squared of π. We want to rearrange this to find an expression for the tan squared of two π₯.

So weβll start by subtracting one from both sides, giving us the tan squared of π is equivalent to the sec squared of π minus one. Then we just need to replace π with two π₯. So now we can replace the tan squared of two π₯ with the sec squared of two π₯ minus one. So by replacing the tan squared of two π₯ in our expression for π¦, we get that π¦ is equal to one-third times the sec squared of two π₯ plus two-thirds times the sec squared of two π₯ minus one. And now we can see why this is useful. We just distribute two-thirds over our parentheses. This gives us the following expression. And then we can combine one-third times the sec squared of two π₯ plus two-thirds times the sec squared of two π₯ to just be the sec squared of two π₯.

So after all of this simplification, we were able to rewrite π¦ as the sec squared of two π₯ minus two-thirds. And this is a much easier expression to differentiate than our original expression. Now all thatβs left to do is differentiate this with respect to π₯. Of course, the derivative of the constant negative two-thirds will be equal to zero. So we only need to differentiate the sec squared of two π₯ with respect to π₯. And we have a few choices of how we want to do this. We could use the product rule, the chain rule, or the general power rule. Any method would work. Weβll do this by using the general power rule.

Recall, this tells us for a differentiable function π of π₯ and 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₯ raised to the power of π minus one. We want to use this to differentiate the sec squared of two π₯. So our exponent π will be equal to two, and our function π of π₯ will be the sec of two 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 sec of two π₯ with respect to π₯. And to do this, we need to recall one of our standard trigonometric derivative results. For any real constant π, the derivative of the sec of ππ₯ with respect to π₯ is equal to π times the tan of ππ₯ multiplied by the sec of ππ₯.

In our case, the value of π is equal to two. So we get π prime of π₯ is equal to two tan of two π₯ multiplied by the sec of two π₯. Weβre now ready to find an expression for dπ¦ by dπ₯. Remember, we showed π¦ is equal to the sec squared of two π₯ minus two-thirds. So dπ¦ by dπ₯ will be the derivative of this with respect to π₯. And as we said, weβre going to differentiate the sec squared of two π₯ by using the general power rule. Itβs equal to π times π prime of π₯ multiplied by π of π₯ raised to the power of π minus one. And of course, we know the derivative of the constant negative two-thirds is equal to zero. So we can just substitute in our value of π is two and our expressions for π of π₯ and π prime of π₯.

This gives us dπ¦ by dπ₯ is equal to two times two tan of two π₯ multiplied by the sec of two π₯ times the sec of two π₯ raised to the power of two minus one. And we can simplify this. First, our exponent of two minus one is equal to one. But then we can see we have the sec of two π₯ multiplied by the sec of two π₯, which we can simplify to give us the sec squared of two π₯. Finally, our coefficient two times two is equal to four. And this gives us our final answer. Therefore, by simplifying our expression for π¦ by using the Pythagorean identity and then applying the general power rule, we were able to show if π¦ is equal to one plus two sin squared of two π₯ all divided by three minus three sin squared of two π₯, then dπ¦ by dπ₯ would be equal to four tan of two π₯ times the sec squared of two π₯.