### Video Transcript

Given that π₯ equals five π‘ minus four times the natural log of π‘ and π¦ equals four π‘ plus five times the natural log of three π‘, find dπ¦ by dπ₯.

Here weβve been given a pair of parametric equations. These are equations for π₯ and π¦ in terms of a third parameter. Thatβs π‘. And weβre looking to find the derivative of π¦ with respect to π₯. And so, we recall that as long as dπ₯ by dπ‘ is not equal to zero, the derivative of π¦ with respect to π₯ is equal to dπ¦ by dπ‘ divided by dπ₯ by dπ‘. And so, it should be quite clear to us that weβre going to need to differentiate each of our respective equations with respect to π‘. And weβll do this term by term.

The first term in our equation for π₯ is five π‘. Now, the derivative of five π‘ with respect to π‘ is five. Next, we recall that the derivative of the natural log of ππ₯, where π is some constant, with respect to π₯ is one over π₯. So, the derivative of the natural log of π‘ is one over π‘. And so, the derivative of negative four times the natural log of π‘ is negative four over π‘.

Letβs repeat this process for dπ¦ by dπ‘. The derivative of four π‘ with respect to π‘ is four. Then, the derivative of the natural log of three π‘ must be one over π‘. So, five times the natural log of three π‘ is five over π‘. dπ¦ by dπ₯ is the quotient of these. Itβs dπ¦ by dπ‘ divided by dπ₯ by dπ‘. So, thatβs four plus five over π‘ divided by five minus four over π‘.

Now, at the moment, this doesnβt look particularly nice, so weβre going to deal with the fractions in our numerator and denominator. Weβre going to multiply through by π‘. Then, our numerator becomes four π‘ plus five, and our denominator becomes five π‘ minus four. And so, weβre done. We found dπ¦ by dπ₯. Itβs four π‘ plus five over five π‘ minus four.