### Video Transcript

Given that π₯ equals five π‘π to the π‘ power and π¦ equals three π‘ plus four sin of π‘, find dπ¦ by dπ₯.

In this case, instead of π¦ being defined as a function of π₯, both π₯ and π¦ have been defined in terms of a third variable π‘. And we call these parametric equations. To find dπ¦ by dπ₯, we remind ourselves of the chain rule, which gives us the equation dπ¦ by dπ‘ is equal to dπ¦ by dπ₯ multiplied by dπ₯ by dπ‘. We can rearrange this for dπ¦ by dπ₯. And we get that dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ over dπ₯ by dπ‘. And of course, this only works providing that dπ₯ by dπ‘, the denominator, is not equal to zero.

So, weβre going to need to find dπ¦ by dπ‘ and dπ₯ by dπ‘. dπ¦ by dπ‘ is the derivative of π¦ equals three π‘ add four sin of π‘ with respect to π‘. We can do this term by term, starting with the term three π‘, which we know differentiates to give us three. Now for the term four sin of π‘, we recall that the derivative of sin of π‘ is cos of π‘ with respect to π‘. So, four sin of π‘ differentiates to give us four cos of π‘. So we have that dπ¦ by dπ‘ is equal to three add four cos of π‘.

Now weβve also got to find dπ₯ by dπ‘. This is going to be the derivative of π₯ equals five π‘π to the π‘ power with respect to π‘. One thing to be aware of here is that we have two functions of π‘ being multiplied together. Itβs five π‘ multiplied by π to the π‘ power. So, to differentiate this, weβve got to use the product rule. The product rule tells us that the derivative of function π multiplied by function π is π prime multiplied by π add π multiplied by π prime. Remember, the notation π prime just means the derivative of the function π.

In this case, the function π of π‘ is equal to five π‘ and the function π of π‘ is equal to π to the π‘ power. So in order to use the product rule, we want to find π prime of π‘ and π prime of π‘, the derivative of π and the derivative of π. Well, we know that the derivative of five π‘ is five. To differentiate the function π, we recall the fact that the derivative of π to the π‘ power is just π to the π‘ power. And this is just a general result of differentiation of the exponential function.

Now we can put what we found into the formula for the product rule. So we find that dπ₯ by dπ‘ is equal to π prime multiplied by π, which is five multiplied by π to the π‘ power add π multiplied by π prime, which is five π‘ multiplied by π to the π‘ power. So by the product rule, dπ₯ by dπ‘ is five π to the π‘ power add five π‘π to the π‘ power. Notice that we can take five π to the π‘ power out as a common factor, which gives us five π to the π‘ power multiplied by one add π‘. Or equivalently, we could write this the other way as five π to the π‘ power multiplied by π‘ add one.

We can now use the fact that dπ¦ by dπ₯ equals dπ¦ by dπ‘ over dπ₯ by dπ‘. Then we have that dπ¦ by dπ₯ equals three add four cos of π‘ over five π to the π‘ multiplied by π‘ add one. And that gives us our final answer.