### Video Transcript

Given that π₯ equals three π to the power of five π‘ and π¦ equals π‘ times π to the power of negative five π‘, find dπ¦ by dπ₯.

In this question, weβve been given a pair of parametric equations. These are equations for π₯ and π¦ in terms of a third parameter; here, thatβs π‘. Weβre asked to differentiate π¦ with respect to π₯. And so we recall that if dπ¦ by dπ‘ is not equal to zero, dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ divided by dπ₯ by dπ‘. And so we can see that weβre going to need to differentiate the equation for π₯ with respect to π‘ and the equation for π¦ with respect to π‘.

Weβll begin by differentiating π₯ with respect to π‘. Now, π₯ is equal to three π to the power of five π‘. And we know that to differentiate π to the power of ππ₯ for some constant value π, we get π times π to the power of ππ₯. Well, in this case, weβre differentiating three π to the power of five π‘. So we multiply the entire term by the coefficient of π‘, which is five. And we see that dπ₯ by dπ‘ is five times three π to the power of five π‘, which is 15π to the power of five π‘.

Next, we need to differentiate π¦ with respect to π‘. But we see that π¦ itself is the product of two differentiable functions. Itβs π‘ times π to the power of negative five π‘. And so weβre going to use the product rule. This says that the derivative of the product of two differentiable functions π’ and π£ is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. Well, our two functions are π‘ and π to the power of negative five π‘. So dπ¦ by dπ‘ is π‘ times the derivative of π to the power of negative five π‘ with respect to π‘ plus π to the power of negative five π‘ times the derivative of π‘ with respect to π‘.

Now, the derivative of π to the power of negative five π‘ we know to be equal to negative five π to the power of negative five π‘. And the derivative of π‘ with respect to π‘ is simply one. So dπ¦ by dπ‘ is π‘ times negative five π to the power of negative five π‘ plus π to the power of negative five π‘, which simplifies as shown. dπ¦ by dπ₯ is the quotient of these. Itβs dπ¦ by dπ‘ divided by dπ₯ by dπ‘. So thatβs negative five π‘π to the power of negative five π‘ plus π to the power of negative five π‘ all over 15π to the power five π‘.

By recognizing that π to the power of negative five π‘ is the same as one over π to the power of five π‘ and then factoring by one over π to the power of five π‘. We see we can write this as one over π to the power of five π‘ times negative five π‘ plus one over 15π to the power of five π‘. But of course, we would generally write this second fraction as one minus five π‘ over 15π to the power of five π‘. Now, when we multiply two numbers with exponents, we add those exponents. So π to the power of five π‘ times 15π to the power of five π‘ is 15π to the power of 10π‘.

And we therefore find dπ¦ by dπ₯ to be equal to one minus five π‘ over 15π to the power of 10π‘.