### Video Transcript

Given that π₯ is equal to the square root of negative π‘ plus five and π¦ is equal to the square root of two π‘ plus one, find dπ¦ by dπ₯ at π‘ equals zero.

Here, we have a pair of parametric equations. These are equations for π₯ and π¦ in terms of a third parameter, π‘. The question is asking us to differentiate the function of π¦ with respect to π₯ and then evaluate it when π‘ equals zero. And so, we recall that, to find dπ¦ by dπ₯ when dealing with a pair of parametric equations, as long as dπ₯ by dπ‘ is not equal to zero, dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ over dπ₯ by dπ‘. And so, it should be quite clear that weβre going to need to differentiate each of our equations with respect to π‘.

dπ₯ by dπ‘ is the derivative of the square root of negative π‘ plus five with respect to π‘. Now, we have got a bit of a problem here. This is a composite function. Itβs a function of a function. And so, weβre going to use the chain rule. This says that if π and π are differentiable functions such that π¦ is equal to π of π’ and π’ is equal to π of π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. Now, of course, our function is π₯ and itβs in terms of π‘. So we say that dπ₯ by dπ‘ is equal to dπ₯ by dπ’ times dπ’ by dπ‘.

So weβre going to let our substitution π’ be equal to negative π‘ plus five such that π₯ is equal to the square root of π’. Weβll rewrite the square root of π’ as π’ to the power of one-half. Then we differentiate each function where the derivative of π’ with respect to π‘ is negative one. Then, to differentiate this power term, we multiply the entire term by the exponent, then reduce the exponent by one. So dπ₯ by dπ’ is a half π’ to the power of negative one-half. dπ₯ by dπ‘ is the product of these. So itβs negative one times a half π’ to the power of negative one-half, which is negative one-half π’ to the power of negative one-half.

Now, of course, weβre actually differentiating π₯ with respect to π‘. So weβre going to need to replace our value for π’ with negative π‘ plus five. And then, we find dπ₯ by dπ‘ to be equal to negative one-half times negative π‘ plus five to the power of negative one-half. Weβre going to repeat this process, this time differentiating π¦ with respect to π‘. So thatβs the derivative of the square root of two π‘ plus one with respect to π‘. Weβre going to change our chain rule up a little bit. And this time, weβre going to let π’ be equal to two π‘ plus one and π¦ is equal to π’ to the power of one-half.

This time, dπ’ by dπ‘ is two but dπ¦ by dπ’ is the same as dπ₯ by dπ’ from before. So dπ¦ by dπ‘ is two times one-half π’ to the power of negative one-half, which is π’ to the power of negative one-half. But remember, we said π’ is equal to two π‘ plus one. So dπ¦ by dπ‘ is two π‘ plus one to the power of negative one-half. dπ¦ by dπ₯ is the quotient of these. Itβs dπ¦ by dπ‘ divided by dπ₯ by dπ‘. So we get two π‘ plus one to the power of negative one-half over negative one-half times negative π‘ plus five to the power of negative one-half.

Now, we know that dividing by a fraction is the same as multiplying by the reciprocal of that fraction. So dividing by negative one-half is the same as multiplying by negative two. We also know that a power of negative one-half is one over the square root. So, essentially, we find the reciprocal. And we get negative two times negative π‘ plus five to the power of one-half over two π‘ plus one to the power of one-half.

Now, weβre looking to evaluate this when π‘ is equal to zero. So letβs substitute π‘ equals zero into this equation. When we do, we get negative two times five to the power of one-half over one to the power of one-half. But a power of one-half is the same as finding the square root. And the square root of one is just one. So this becomes negative two root five. And so, given our pair of parametric equations, we found the value of dπ¦ by dπ₯ when π‘ is equal to zero. Itβs negative two root five.