### Video Transcript

If π¦ is equal to the cos of six π₯ all divided by one minus the sin of six π₯, find the derivative of π¦ with respect to π₯.

We need to find the derivative of π¦ with respect to π₯. And we can see that π¦ is given as the quotient of two functions; itβs the cos of six π₯ divided by one minus the sin of six π₯. And since π¦ is the quotient of two functions and we know how to differentiate both the numerator and denominator with respect to π₯, we know we can use the quotient rule. However, itβs always worth checking thereβs not a simpler method. For example, we could try using the double-angle formulas to simplify this expression. Or we could write this as the cos of six π₯ multiplied by one over one minus the sin of six π₯ and then apply the product rule and the general power rule.

However, in this case, these methods are not any easier than just applying the quotient rule. So, weβd evaluate this by using the quotient rule. Letβs start by recalling what we mean by the quotient rule. If π¦ is equal to the quotient of two differentiable functions π of π₯ over π of π₯, then the derivative 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. And, of course, we know we canβt divide by zero as this will only be valid when π of π₯ is not equal to zero. So, letβs start by setting π of π₯ to be the function in our numerator β thatβs the cos of six π₯ β and π of π₯ to be the function in our denominator; thatβs one minus the sin of six π₯.

We can see that both π of π₯ and π of π₯ are differentiable. In fact, we know how to evaluate their derivatives. 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₯. Thatβs the derivative of the cos of six π₯ with respect to π₯. And we can evaluate this by recalling one of our standard trigonometric derivative results. We know for any real constant π, the derivative of the cos of ππ₯ with respect to π₯ is equal to negative π times the sin of ππ₯. In this case, the coefficient of π₯ is equal to six. So, weβll use π is equal to six, giving us that π prime of π₯ is equal to negative six times the sin of six π₯.

We now need to find an expression for π prime of π₯. Thatβs the derivative of one minus the sin of six π₯ with respect to π₯. Once again, we can evaluate this by recalling one of our standard trigonometric derivative results. We know for any real constant π, the derivative of the sin of ππ₯ with respect to π₯ is equal to π times the cos of π₯. Weβll need to evaluate this derivative term by term. First, the derivative of the constant one is equal to zero. Now, we need to subtract the derivative of the sin of six π₯ with respect to π₯. And of course, we can evaluate this derivative by using our standard trigonometric derivative result with π equal to six. We get six times the cos of six π₯. And of course, this just simplifies to give us negative six times the cos of six π₯.

Now that we found expressions for π prime of π₯ and π prime of π₯, weβre ready to find an expression for dπ¦ by dπ₯ by using the quotient rule. Itβs 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. Substituting in our expressions for π of π₯, π of π₯, π prime of π₯, and π prime of π₯, we get that dπ¦ by dπ₯ is equal to negative six times the sin of six π₯ multiplied by one minus the sin of six π₯ minus negative six times the cos of six π₯ multiplied by the cos of six π₯ all divided by one minus the sin of six π₯ all squared.

And now we can start simplifying. First, the second term in our numerator can be simplified to give us six times the cos squared of six π₯. We can also simplify the first term in our numerator by distributing over our parentheses. Doing this gives us negative six times the sin of six π₯ plus six times the sin squared of six π₯ plus six times the cos squared of six π₯ all divided by one minus the sin of six π₯ all squared. And we can simplify even further. First, we can take out the shared factor of six in all three terms in our numerator. This gives us a new numerator of six times negative the sin of six π₯ plus the sin squared of six π₯ plus the cos squared of six π₯.

And now, we can notice something interesting. We can in fact simplify this expression even further by using 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 equivalent to one. This would even be true if we replace π with six π₯. This means by using the Pythagorean identity, we can simplify this expression to give us six times negative the sin of six π₯ plus one all divided by one minus the sin of six π₯ all squared. And now, thereβs one more piece of simplification weβll do. To see this, we need to rearrange the two terms inside the parentheses in our numerator.

Doing this, we get six times one minus the sin of six π₯ all divided by one minus the sin of six π₯ all squared. And now, we can see we can cancel the shared factor of one minus the sin of six π₯ in our numerator and our denominator. And this leaves us with our final answer, six divided by one minus the sin of six π₯.

Therefore, by using the quotient rule, we were able to show if π¦ is equal to the cos of six π₯ all divided by one minus the sin of six π₯, then the derivative of π¦ with respect to π₯ is equal to six divided by one minus the sin of six π₯.