### Video Transcript

Find dπ¦ by dπ₯ for the function π¦ is equal to negative three times the csc of four π₯ to the fifth power minus three.

The question wants us to find the first derivative of π¦ with respect to π₯. And we can see that π¦ is the composition of functions; itβs a composition of the cosecant function and the polynomial four π₯ to the fifth power minus three. And we know how to find the derivative of the composition of two functions by using the chain rule.

We recall the chain rule tells us, if we have π¦ is a function of π’ and π’ in turn is a function of π₯, then the derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to π’ multiplied by the derivative of π’ with respect to π₯. So, if we set π’ to be our inner function, thatβs the polynomial four π₯ to the fifth power minus three, then this tells us that π¦ is equal to negative three times the csc of π’. Weβve rewritten π¦ to be a function of π’, and π’, in turn, is a function of π₯.

So, we can evaluate this derivative by using the chain rule. We get dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. And we can actually evaluate both of these derivatives. First, dπ¦ by dπ’ is the derivative of π¦ with respect to π’ and π¦ is equal to negative three times the csc of π’. So, dπ¦ by dπ’ is the derivative of negative three times the csc of π’ with respect to π’.

Next, dπ’ by dπ₯ is the derivative of π’ with respect to π₯, and π’ is four π₯ to the fifth power minus three. So, dπ’ by dπ₯ is equal to the derivative of four π₯ to the fifth power minus three with respect to π₯. And we can just evaluate both of these derivatives. First, we know for any constant π, the derivative of π times the csc of π’ with respect to π’ is equal to negative π times the cot of π’ times the csc of π’. So, applying this to our first derivative, we get negative one multiplied by negative three cot of π’ times the csc of π’.

And our second derivative is just the derivative of a polynomial. We can do this by using the power rule for differentiation. We multiply by our exponent of π₯ and reduce this exponent by one. This gives us 20π₯ to the fourth power. So, we found the following expression for dπ¦ by dπ₯. And we can simplify this expression slightly. First, negative one multiplied by negative three is equal to three. Next, three multiplied by 20π₯ to the fourth power is equal to 60π₯ to the fourth power.

This gives us 60π₯ to the fourth power times the cot of π’ times the csc of π’. And we could leave our answer like this. However, remember, weβre trying to find an expression for the derivative of π¦ with respect to π₯. So, we want our answer to be in terms of π₯. And to do this, weβll use our substitution π’ is equal to four π₯ to the fifth power minus three. And using this substitution, we get our final answer.

If π¦ is equal to negative three times the csc of four π₯ to the fifth power minus three, then dπ¦ by dπ₯ is equal to 60π₯ to the fourth power times the cot of four π₯ to the fifth power minus three multiplied by the csc of four π₯ to the fifth power minus three.