Video Transcript
Find the derivative of the function 𝑦 is equal to the cot squared of the sin of 𝜃.
The question wants us to find the derivative of our function 𝑦. And we can see that 𝑦 is given in terms of 𝜃. So, we want to find the derivative of 𝑦 with respect to 𝜃. There’s a few different approaches we could use to differentiate this function. First, we notice that we’re taking the cot squared of the sin of 𝜃. So, instead of writing 𝑦 as a square, we could write 𝑦 as the cot of the sin of 𝜃 multiplied by itself. And since 𝑦 is then the product of two functions, we could try and differentiate this by using the product rule.
And this would work. However, we’re going to do this by using the chain rule twice. So, to use the chain rule, first we’re going to consider our function 𝑦 as the composition of two functions. First, it’s the cot of the sin of 𝜃. And then, we square this value. We now recall the following version of the chain rule. If we have 𝑦 is a function of 𝑢 and 𝑢 in turn is a function of 𝜃, then d𝑦 by d𝜃 is equal to d𝑦 by d𝑢 times d𝑢 by d𝜃.
So, to use the chain rule to find d𝑦 by d𝜃, we’ll set our functions 𝑢 to be our inner function of the cot of the sin of 𝜃. Using this, we can see we’ve now rewritten 𝑦 to be equal to 𝑢 squared. 𝑦 is a function of 𝑢, and 𝑢 in turn is a function of 𝜃. So, by using the chain rule, we now have d𝑦 by d𝜃 is equal to d𝑦 by d𝑢 times d𝑢 by d𝜃. At this point, we can find d𝑦 by d𝑢. However, we can’t directly evaluate d𝑢 by d𝜃. This is because 𝑢 is the composition of two functions; it’s the cot of the sin of 𝜃.
So, to find the derivative of 𝑢 with respect to 𝜃, we’re going to need to apply the chain rule again. This time, we’ll write the chain rule slightly differently. If we have 𝑢 is a function of 𝑣 and 𝑣 in turn is a function of 𝜃, then d𝑢 by d𝜃 is equal to d𝑢 by d𝑣 times d𝑣 by d𝜃. This time, to use the chain rule, we’ll set our function 𝑣 to be our inner function the sin of 𝜃. And using this, we’ve now rewritten 𝑢 to be the cot of 𝑣. 𝑢 is a function of 𝑣, and 𝑣 in turn is a function of 𝜃.
So, now, we can use our chain rule to find an expression for d𝑢 by d𝜃. And using this, we have d𝑢 by d𝜃 is equal to d𝑢 by d𝑣 times d𝑣 by d𝜃. So, we now have this new expression for d𝑦 by d𝜃. And we can actually evaluate all parts of this expression. So, let’s evaluate each of these parts separately.
First, d𝑦 by d𝑢 is the derivative of 𝑦 with respect to 𝑢. We know that 𝑦 is equal to 𝑢 squared, so this is the derivative of 𝑢 squared with respect to 𝑢. And, of course, we can evaluate this derivative by using the power rule for differentiation. We multiply by the exponent of 𝑢 and reduce this exponent by one. This gives us two 𝑢.
Let’s now evaluate our second part, the derivative of 𝑢 with respect to 𝑣. We know that 𝑢 is equal to the cot of 𝑣. So, we need to differentiate the cot of 𝑣 with respect to 𝑣. And this is one of our standard derivative results for trigonometric functions. We know the derivative of the cot of 𝑥 with respect to 𝑥 is equal to negative the csc squared of 𝑥. So, the derivative of the cot of 𝑣 with respect to 𝑣 is negative the csc squared of 𝑣.
Finally, we want to find an expression for d𝑣 by d𝜃. We know that 𝑣 is equal to the sin of 𝜃. So, this is the derivative of the sin of 𝜃 with respect to 𝜃. And again, this is one of our standard trigonometric derivative results which we should commit to memory. The derivative of the sin of 𝜃 with respect to 𝜃 is equal to the cos of 𝜃. So, we’ve now shown that d𝑦 by d𝜃 is equal to two 𝑢 times negative the csc squared of 𝑣 multiplied by the cos of 𝜃.
Multiplying these together and rearranging, we get negative two cos of 𝜃 times 𝑢 multiplied by the csc squared of 𝑣. And we could leave our answer like this. However, remember, we’re trying to find an expression for the derivative of 𝑦 with respect to 𝜃. So, we should give our answer in terms of 𝜃. And we can do this by remembering that 𝑢 is equal to the cot of the sin of 𝜃 and 𝑣 is equal to the sin of 𝜃.
So, substituting in these expressions for 𝑢 and 𝑣, we get that d𝑦 by d𝜃 is equal to negative two cos of 𝜃 times the cot of the sin of 𝜃 multiplied by the csc squared of the sin of 𝜃. And this is our final answer. Therefore, by using the chain rule twice, we were able to show the derivative of the function 𝑦 is equal to the cot squared of the sin of 𝜃. Is equal to negative two cos of 𝜃 times the cot of the sin of 𝜃 multiplied by the csc squared of the sin of 𝜃.
