### Video Transcript

If π₯ equals π‘ squared plus four π‘ minus two and π¦ equals two π‘ squared minus three, find dπ¦ dπ₯ at π‘ equals one.

Here we have two equations in terms of π‘. And we are asked to find dπ¦ dπ₯ at π‘ equals one. So once we find dπ¦ dπ₯, we will plug in one. So if this is our equation for π₯, we can take it and find the derivative of π₯ with respect to π‘, dπ₯ dπ‘. And if this is our equation for π¦, we can find dπ¦ dπ‘, derivative of π¦ with respect to π‘.

Now both of these are helpful because if we would take dπ¦ dπ‘ and divide it by dπ₯ dπ‘, this is the same as taking dπ¦ dπ‘ and multiplying by the reciprocal for denominator, dπ‘ dπ₯. The dπ‘s cancel, and weβre left with dπ¦ dπ₯, which is exactly what weβre supposed to find, and then plug in one.

So letβs begin by finding dπ₯ dπ‘, the derivative of π₯ with respect to π‘. So we need to use the power rule. π‘ squared will turn into two π‘ or two π‘ to the first power, because we bring the exponent down to the front, the power down to the front, and then subtract one. And two minus one is one. So now for four π‘, our power would be one. The exponent would be one. We would bring that down to be with the four. And one times four would give us four. And then the power of π‘ was one. So we take one and subtract one, which would be zero. And π‘ to the zero power, actually anything to the zero power is simply one. And four times one is one. And then the derivative, the constant, is just zero. So we donβt have to do anything with the minus two.

So now letβs find dπ¦ dπ‘. So we take two π‘ squared. The power of two goes down in front to be with the two. And two times two is four. And then our exponent will turn into two minus one, so π‘ to the one power, which is just π‘. Now we have a constant. We have minus three. So its derivative is simply zero. So as we said before, to find dπ¦ dπ₯, we take dπ¦ dπ‘, which is four π‘, and divide it by dπ₯ dπ‘, two π‘ plus four.

Now we are able to take a factor of two out of the numerator and denominator, making dπ¦ dπ₯ equal to two π‘ over π‘ plus two. So now that we have dπ¦ dπ₯, weβre asked to find that at π‘ equals one. So if we would plug in one for π‘, weβll have two times one over one plus two, resulting in two-thirds. Therefore, dπ¦ dπ₯ at π‘ equals one is equal to two over three or two-thirds.