Video: Finding the First Derivative of the Quotient of Trigonometric Functions Using the Quotient Rule

If 𝑦 = sin π‘₯/(1 βˆ’ cos π‘₯), which of the following is the same as 𝑦′?[A] 𝑦 [B] 2𝑦 csc π‘₯ [C] 𝑦 csc π‘₯ [D] βˆ’π‘¦ csc π‘₯

07:16

Video Transcript

If 𝑦 is equal to the sin of π‘₯ divided by one minus the cos of π‘₯, which of the following is the same as 𝑦 prime? Option (A) 𝑦, option (B) two 𝑦 times the csc of π‘₯, option (C) 𝑦 times the csc of π‘₯, or option (D) negative 𝑦 times the csc of π‘₯.

We’re given four options. We need to determine which of these four options is the same as 𝑦 prime given that 𝑦 is equal to the sin of π‘₯ divided by one minus the cos of π‘₯. To do this, we need to recall what we mean by 𝑦 prime. Since 𝑦 is a function in π‘₯, this will be the derivative of 𝑦 with respect to π‘₯. So we’re going to need to differentiate this expression for 𝑦 with respect to π‘₯. And there’s several different ways of doing this. For example, we could write this as the sin of π‘₯ multiplied by one over one minus the cos of π‘₯ and then use the product rule and the general power rule. And this would work. However, we’re going to do this by using the quotient rule.

So let’s start by recalling what we mean by the quotient rule. If we’re trying to differentiate the quotient of two differentiable functions 𝑓 of π‘₯ over 𝑔 of π‘₯, then this is equal to 𝑓 prime of π‘₯ times 𝑔 of π‘₯ minus 𝑔 prime of π‘₯ times 𝑓 of π‘₯ all divided by 𝑔 of π‘₯ all squared. And we know since we can’t divide by zero, this will only be valid when 𝑔 of π‘₯ is not equal to zero. So to apply the quotient rule to this function, we’ll set 𝑓 of π‘₯ to be the function in our numerator, that’s the sin of π‘₯, and 𝑔 of π‘₯ to be the function in our denominator, that’s one minus the cos of π‘₯.

And it’s worth pointing out both of these are differentiable. In fact, we know how to differentiate them. So to use the quotient rule, we need to find expressions for 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. Let’s start with 𝑓 prime of π‘₯. That’s the derivative of the sin of π‘₯ with respect to π‘₯. And in fact, this is a standard trigonometric derivative result which we should commit to memory. The derivative of the sin of π‘₯ with respect to π‘₯ is equal to the cos of π‘₯. So by using this, we have 𝑓 prime of π‘₯ is equal to the cos of π‘₯.

Let’s now find an expression for 𝑔 prime of π‘₯. That’s the derivative of one minus the cos of π‘₯ with respect to π‘₯. Once again, to do this, we need to recall one of our standard trigonometric derivative results which we should commit to memory. The derivative of the cos of π‘₯ with respect to π‘₯ is equal to negative the sin of π‘₯ However, in this case, there’s a few more steps. First, we need to evaluate this derivative term by term. We’ll start by differentiating the constant one with respect to π‘₯. Of course, this is a constant, so its derivative is equal to zero.

Now, we need to subtract the derivative of the cos of π‘₯. Of course, we’ve already shown this is equal to negative the sin of π‘₯. And now, we can simplify this. We have negative one times negative the sin of π‘₯ is just equal to the sin of π‘₯. So we’ve now shown that 𝑔 prime of π‘₯ is equal to the sin of π‘₯. We’re now ready to find an expression for 𝑦 prime by using the quotient rule. It’s equal to 𝑓 prime of π‘₯ times 𝑔 of π‘₯ minus 𝑔 prime of π‘₯ times 𝑓 of π‘₯ all divided by 𝑔 of π‘₯ all squared. Substituting in our expressions for 𝑓 of π‘₯ , 𝑔 of π‘₯, 𝑓 prime of π‘₯, and 𝑔 prime of π‘₯, we can show that 𝑦 prime is equal to the cos of π‘₯ times one minus the cos of π‘₯ minus the sin of π‘₯ times the sin of π‘₯ all divided by one minus the cos of π‘₯ all squared.

And now, we have an expression for 𝑦 prime. So there’s several different things we could attempt to do at this stage of the question. For example, we could start finding expressions for each of our four options and see if any of these are equal to our expression for 𝑦 prime. However, it’s often easiest to start by simplifying our expression for 𝑦 prime. So that’s what we’ll do. In our numerator, we’ll distribute the cos of π‘₯ over the parentheses in our first term. And we’ll simplify the second term in our numerator to be the sin squared of π‘₯. Doing this, we get the cos of π‘₯ minus the cos squared of π‘₯ minus the sin squared of π‘₯ all divided by one minus the cos of π‘₯ all squared.

And at this point, it can be difficult to see how we could simplify this expression any further. To notice this, we need to see that the cos squared of π‘₯ and the sin squared of π‘₯ both share a factor of negative one. Taking out the shared factor of negative one, we get the cos of π‘₯ minus the cos squared of π‘₯ plus the sin squared of π‘₯ all divided by one minus the cos of π‘₯ all squared. And now, we should be able to notice something interesting about the cos squared of π‘₯ plus the sin squared of π‘₯. We can in fact simplify this by using the Pythagorean identity.

We recall this tells us for any value of πœƒ, the sin squared of πœƒ plus the cos squared of πœƒ is equal to one. So if, instead of πœƒ, we had π‘₯ and we rearrange these two terms, we can see the cos squared of π‘₯ plus the sin squared of π‘₯ is also equivalent to one. So by using this, we get the cos of π‘₯ minus one all divided by one minus the cos of π‘₯ all squared. And in fact, there’s one more piece of simplification we can do. We need to notice our numerator is in fact a linear multiple of what’s inside the parentheses in our denominator. In other words, we can rewrite our numerator as negative one multiplied by one minus the cos of π‘₯.

And then by rewriting our numerator in this fashion, we can cancel one of the shared factors of one minus the cos of π‘₯ from our numerator and our denominator. This leaves us with negative one divided by one minus the cos of π‘₯. We can now see which of our four options this can be equal to. Let’s start with option (A). We can check if this is equal to 𝑦. We know that 𝑦 is equal to the sin of π‘₯ divided by one minus the cos of π‘₯. We can see our answer is almost the same as this. However, in our numerator, we have negative one and not the sin of π‘₯. So our answer can’t be option (A).

And we could check options (B), (C), and (D) using exactly the same method. However, we can notice something interesting about all four of these options. They all have a factor of 𝑦. So we should try rewriting our answer to have a factor of 𝑦. And when we were checking option (A), we did see this was almost the same. We just needed a factor of the sin of π‘₯ in our numerator. One way of getting a factor of the sin of π‘₯ in our numerator is to multiply both the numerator and the denominator by the sin of π‘₯. However, we can also see all three of our remaining options contain the csc of π‘₯.

So instead of dividing by the sin of π‘₯, we can instead multiply by the csc of π‘₯. And of course, we can do this because we know one over the sin of π‘₯ is equivalent to the csc of π‘₯. This means we can rewrite 𝑦 prime as negative one over one minus the cos of π‘₯ multiplied by the sin of π‘₯ multiplied by the csc of π‘₯. If we then move just the factor of the sin of π‘₯ into our numerator, we get negative one times the sin of π‘₯ divided by one minus the cos of π‘₯ all multiplied by the csc of π‘₯. And of course, we know the sin of π‘₯ divided by one minus the cos of π‘₯ is equal to 𝑦.

So in fact, we’ve managed to rewrite our expression for 𝑦 prime as negative 𝑦 times the csc of π‘₯ which we can see is our option (D). And of course, we can also see this means it can’t be option (C) or option (B). Option (C) is off by a factor of negative one and option (B) is off by a factor of negative two. Therefore, by using the quotient rule and some of our trigonometric identities, we were able to show if 𝑦 is equal to the sin of π‘₯ divided by one minus the cos of π‘₯, then 𝑦 prime is equal to negative 𝑦 times the csc of π‘₯ which is equal to our option (D).

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