Video Transcript
Find the derivative of the csc of two 𝑥 plus the cot of nine 𝑥 multiplied by the csc of two 𝑥 minus the cot of nine 𝑥 with respect to 𝑥.
We’re asked to find the derivative of the product of two trigonometric functions. So, we could do this by using the product rule, and this would work and give us the correct answer. However, we can also multiply our two products together and notice this is a factoring of a difference between squares. So we’ll start by multiplying these two together. This means we now need to evaluate the derivative of the csc squared of two 𝑥 minus the cot squared of nine 𝑥 with respect to 𝑥. And we can evaluate this derivative term by term. In fact, there’s several different methods we could use to do this.
For example, we could use the product rule, the chain rule, or the general power rule. It doesn’t matter which you would prefer; it’s all personal preference. We’re going to do this by using the general power rule. We recall this tells us for constant 𝑛 and differentiable function 𝑓 of 𝑥, the derivative of 𝑓 of 𝑥 all raised to the 𝑛th power with respect to 𝑥 is equal to 𝑛 times 𝑓 prime of 𝑥 multiplied by 𝑓 of 𝑥 raised to the power of 𝑛 minus one. And we need to apply this to each term separately. Let’s start with our first term.
We can see our exponent 𝑛 will be equal to two, and our inner function 𝑓 of 𝑥 will be the csc of two 𝑥. And to use the general power rule, we’re going to need to find an expression for 𝑓 prime of 𝑥. That’s the derivative of the csc of two 𝑥 with respect to 𝑥. And to evaluate this derivative, we need to recall one of our standard trigonometric derivative results. For any real constant 𝑎, the derivative of the csc of 𝑎𝑥 with respect to 𝑥 is equal to negative 𝑎 times the cot of 𝑎𝑥 multiplied by the csc of 𝑎𝑥. So, by setting our value of 𝑎 equal to two, we get that 𝑓 prime of 𝑥 is equal to negative two cot of two 𝑥 multiplied by the csc of two 𝑥.
Now that we’ve found an expression for 𝑓 prime of 𝑥, we can use the general power rule to differentiate our first term. Substituting in our value of 𝑛 is equal to two and our expressions for 𝑓 of 𝑥 and 𝑓 prime of 𝑥, we get the derivative of the csc squared of two 𝑥 with respect to 𝑥 is equal to two times negative two cot of two 𝑥 multiplied by the csc of two 𝑥 times the csc of two 𝑥 all raised to the power of two minus one. And we can simplify this. First, we have two multiplied by negative two is equal to negative four. Next, in our exponent, we have two minus one, which is equal to one. But then, we have the csc of two 𝑥 multiplied by the csc of two 𝑥, which is the csc squared of two 𝑥. So, this expression simplified to give us negative four cot of two 𝑥 multiplied by the csc squared of two 𝑥.
We now want to do exactly the same to evaluate the derivative of our second term. Once again, we see our exponent 𝑛 will be two. This time, we call our inner function 𝑔 of 𝑥 which is the cot of nine 𝑥. Once again, to apply the general power rule, we’re going to need to find an expression for 𝑔 prime of 𝑥. That’s the derivative of the cot of nine 𝑥 with respect to 𝑥. And we can evaluate this by using one of our standard trigonometric derivative results. For any real constant 𝑎, the derivative of the cot of 𝑎𝑥 with respect to 𝑥 is equal to negative 𝑎 times the csc squared of 𝑎𝑥. So by setting our value of 𝑎 equal to nine, we get 𝑔 prime of 𝑥 is equal to negative nine times the csc squared of nine 𝑥.
And now that we’ve found an expression for 𝑔 prime of 𝑥, we can use the general power rule to find the derivative of our second term. Using 𝑛 is equal to two and our expressions for 𝑔 of 𝑥 and 𝑔 prime of 𝑥, we now need to subtract two times negative nine csc squared of nine 𝑥 multiplied by the cot of nine 𝑥 raised to the power of two minus one. And we can simplify this expression slightly. First, negative two multiplied by negative nine is equal to positive 18. And we can simplify our exponent two minus one is equal to one.
This gives us the following expression for our derivative, and we could leave our answer like this. However, we could also simplify this even more. And to do this, we’re going to recall the following trigonometric identity. The cot of 𝜃 is equivalent to the cos of 𝜃 divided by the sin of 𝜃. This means we can rewrite the cot of two 𝑥 as the cos of two 𝑥 divided by the sin of two 𝑥 and the cot of nine 𝑥 as the cos of nine 𝑥 divided by the sin of nine 𝑥. So by using this identity, we can rewrite our answer in the following way.
And there’s one more piece of simplification we need to do. We need to recall one final trigonometric identity. The csc of 𝜃 is equivalent to one over the sin of 𝜃. This means instead of dividing by the sin of two 𝑥, we can instead multiply by the csc of two 𝑥. And instead of dividing by the sin of nine 𝑥, we can multiply by the csc of nine 𝑥. And we can do this by raising our exponents up by one. Finally, after some rearranging, we get our final answer.
Therefore, we were able to show the derivative of the csc of two 𝑥 plus the cot of nine 𝑥 multiplied by the csc of two 𝑥 minus the cot of two 𝑥 with respect to 𝑥 is equal to negative four cos of two 𝑥 times the csc cubed of two 𝑥 plus 18 cos of nine 𝑥 multiplied by the csc cubed of nine 𝑥.
