Question Video: Differentiating Functions Involving Trigonometric Ratios Using the Difference of Two Squares and the Chain Rule

If 𝑦 = (csc π‘₯ + cot 8π‘₯)(csc π‘₯ βˆ’ cot 8π‘₯), find 𝑦′.

06:40

Video Transcript

If 𝑦 is equal to the csc of π‘₯ plus the cot of eight π‘₯ multiplied by the csc of π‘₯ minus the cot of eight π‘₯, find 𝑦 prime.

We’re given 𝑦 which is a function in π‘₯. We need to find an expression for 𝑦 prime. That’s the derivative of 𝑦 with respect to π‘₯. And there are several different methods we could use. For example, we could notice that 𝑦 is written as the product of two functions. And we know how to differentiate each of these functions individually, so we could differentiate 𝑦 by using the product rule. And this would work and give us the correct answer. However, we could also distribute over our parentheses or use a difference between squares to simplify this equation.

We know that π‘Ž plus 𝑏 multiplied by π‘Ž minus 𝑏 is equal to π‘Ž squared minus 𝑏 squared, so we can use this to simplify our expression for 𝑦. We get that 𝑦 is equal to the csc squared of π‘₯ minus the cot squared of eight π‘₯. And now we have a couple of options of how to differentiate each of these expressions. For example, we could use the product rule to evaluate each of these derivatives. However, since both of these are raised to the power of two, we’re going to do this by using the general power rule.

Let’s start by recalling the general power rule. The general power rule tells us for any differentiable function 𝑔 of π‘₯ and real constant 𝑛, the derivative of 𝑔 of π‘₯ all raised to the power of 𝑛 with respect to π‘₯ is equal to 𝑛 times 𝑔 prime of π‘₯ multiplied by 𝑔 of π‘₯ all raised to the power of 𝑛 minus one. We’re now ready to start finding our expression for 𝑦 prime. That’s the derivative of the csc squared of π‘₯ minus the cot squared of eight π‘₯ with respect to π‘₯. And we’ll evaluate this term by term. We need to find the derivative of the csc squared of π‘₯ with respect to π‘₯ minus the derivative of the cot squared of eight π‘₯ with respect to π‘₯. And we’ll evaluate both of these by using the general power rule.

Let’s start with our first derivative. We’ll set 𝑔 of π‘₯ to be the csc of π‘₯ and our exponent 𝑛 equal to two. Now, to use the general power rule with 𝑔 of π‘₯ equal to the csc of π‘₯ and 𝑛 equal to two, we need to find an expression for 𝑔 prime of π‘₯. That’s the derivative of the csc of π‘₯ with respect to π‘₯. And this is a standard trigonometric derivative result we should commit to memory. The derivative of the csc of π‘₯ with respect to π‘₯ is equal to negative the csc of π‘₯ times the cot of π‘₯. So we’ve now shown 𝑔 prime of π‘₯ is equal to negative the csc of π‘₯ times the cot of π‘₯.

We’re now ready to use the general power rule to help us find an expression for the derivative of the csc squared of π‘₯ with respect to π‘₯. It’s equal to two times negative the csc of π‘₯ times the cot of π‘₯ multiplied by the csc of π‘₯ raised to the power of two minus one. And now we can start simplifying. First, in our exponent, two minus one is equal to one and raising our function to the power of one doesn’t change its value. Next, we can simplify the csc of π‘₯ multiplied by the csc of π‘₯ to give us the csc squared of π‘₯. So this gives us negative two csc squared of π‘₯ multiplied by the cot of π‘₯. And we can substitute this directly into our expression for 𝑦 prime.

Now, we need to evaluate the derivative of the cot squared of eight π‘₯ with respect to π‘₯ by using the general power rule. So let’s clear some space so we can evaluate this using the general power rule. First, we’ll set 𝑔 of π‘₯ to be the cot of eight π‘₯ and our exponent 𝑛 equal to two. Next, to use the general power rule, we’re going to need to find an expression for 𝑔 prime of π‘₯. That’s the derivative of the cot of eight π‘₯ with respect to π‘₯. And once again, this is a standard trigonometric derivative result we should commit to memory. For any real constant π‘Ž, the derivative of the cot of π‘Žπ‘₯ with respect to π‘₯ is equal to negative π‘Ž times the csc squared of π‘Žπ‘₯.

In this case, our value of π‘Ž is equal to eight. So we get 𝑔 prime of π‘₯ is equal to negative eight times the csc squared of eight π‘₯. We’re now ready to find an expression for the derivative of the csc squared of eight π‘₯ with respect to π‘₯ by using the general power rule. Our value of 𝑛 is two, 𝑔 of π‘₯ is the cot of eight π‘₯, and we’ve shown 𝑔 prime of π‘₯ is negative eight times the csc squared of eight π‘₯. So substituting these values into our formula, we get two times negative eight csc squared of eight π‘₯ multiplied by the cot of eight π‘₯ all raised to the power of two minus one.

Once again, in our exponent, we’re raising to the power of two minus one, so this doesn’t change the value of our function. And we can also simplify our coefficient. Two multiplied by negative eight is equal to negative 16. This gives us negative 16 csc squared of eight π‘₯ multiplied by the cot of eight π‘₯. Now, just as we did before, we’ll substitute this into our expression for 𝑦 prime. Doing this, we get 𝑦 prime is equal to negative two times the csc squared of π‘₯ multiplied by the cot of π‘₯ minus negative 16 csc squared of eight π‘₯ multiplied by the cot of eight π‘₯.

And we can simplify this. We have negative one multiplied by negative 16, which we know is equal to positive 16. Now, we could leave our answer like this. However, there’s one more piece of simplification we’re going to do. First, we need to recall the trigonometric identity which tells us the cot of πœƒ is equivalent to the cos of πœƒ divided by the sin of πœƒ. Then, we need to recall dividing by the sin of πœƒ is equivalent to multiplying by the csc of πœƒ.

So let’s clear some space and use this identity to simplify our answer. We’re going to use this to turn the cot of π‘₯ into the cos of π‘₯ multiplied by the csc of π‘₯ and the cot of eight π‘₯ into the cos of eight π‘₯ multiplied by the csc of eight π‘₯. Using this identity, we’ve rewritten 𝑦 prime to be equal to negative two times the csc squared of π‘₯ multiplied by the cos of π‘₯ times the csc of π‘₯ plus 16 times the csc squared of eight π‘₯ multiplied by the cos of eight π‘₯ times the csc of eight π‘₯.

And now we just need to simplify this expression. In our first term, we’ve the csc squared of π‘₯ multiplied by the csc of π‘₯. We can simplify this to give us the csc cubed of π‘₯. And we can do something similar in our second term. We have the csc squared of eight π‘₯ multiplied by the csc of eight π‘₯. This is equal to the csc cubed of eight π‘₯. And this gives us our final answer.

Therefore, we were able to show if 𝑦 is equal to the csc of π‘₯ plus the ctn of eight π‘₯ multiplied by the csc of π‘₯ minus the cot of eight π‘₯, then we could simplify this expression by using difference between squares. And then we could evaluate the derivative of this expression with respect to π‘₯ by using the general power rule. We got that 𝑦 prime is equal to negative two cos of π‘₯ multiplied by the csc cubed of π‘₯ plus 16 times the cos of eight π‘₯ multiplied by the csc cubed of eight π‘₯.

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