### Video Transcript

If π₯ equals negative eight π‘ to the fifth power minus eight and π¦ equals the fifth root of π‘ to the sixth power, find dπ¦ by dπ₯ at π‘ equals one.

These functions have been defined parametrically. This means that rather than π¦ being given explicitly as a function of π₯, both π¦ and π₯ are given as functions of a third parameter π‘. Weβre going to find dπ¦ by dπ₯ at the point π‘ equals one. The chain rule tells us that dπ¦ by dπ‘ is equal to dπ¦ by dπ₯ multiplied by dπ₯ by dπ‘. And we can rearrange this to find that dπ¦ by dπ₯ equals dπ¦ by dπ‘ over dπ₯ by dπ‘. And this is what weβre going to use to find dπ¦ by dπ₯.

And of course, this only works providing that the denominator dπ₯ by dπ‘ is not equal to zero. So we need to find dπ¦ by dπ‘ and dπ₯ by dπ‘. Letβs begin by calculating dπ¦ by dπ‘. We have that π¦ equals the fifth root of π‘ to the sixth power. To find dπ¦ by dπ‘, weβre going to differentiate this with respect to π‘. This appears to be a fairly complicated function to differentiate. However, if we remember the fact that the πth root of π₯ to the πth power can be written as π₯ to the power of π over π, we can see that we can rewrite π¦ as π‘ to the six-fifths power.

Then to differentiate this, we recall the power rule of differentiation. This tells us that the derivative of π₯ to the πth power with respect to π₯ is equal to π multiplied by π₯ to the π minus one power. So essentially, we multiply by the power and then subtract one from the power. And applying this to the function π¦, we find that dπ¦ by dπ‘ is equal to six-fifths π‘ raised to the one-fifth power. We find dπ₯ by dπ‘ by differentiating π₯ equals negative π‘ to the fifth power minus eight with respect to π‘.

We can do this term by term starting with the term negative eight π‘ to the fifth power. Applying the power rule to negative eight π‘ to the fifth power, we find that dπ₯ by dπ‘ is equal to negative 40π‘ to the fourth power. As for the negative eight, because this is just a constant, constants differentiate to zero. So dπ₯ by dπ‘ is just negative 40π‘ to the fourth power. Now, remember that we said that dπ₯ by dπ‘ cannot be equal to zero. And because weβre evaluating dπ¦ by dπ₯ at the point π‘ equals one, at this point, dπ₯ by dπ‘ is not equal to zero. So this is okay.

Putting this together, we have that dπ¦ by dπ₯ equals dπ¦ by dπ‘, which is six-fifths π‘ to the one-fifth power over dπ₯ by dπ‘, which is negative 40π‘ to the fourth power. Remember, weβre evaluating this at the point π‘ equals one. So we substitute π‘ equals one, and we remember that raising one to any power is always going to give us one. So both one to the one-fifth power and one to the fourth power are both going to give us one. So weβre left with dπ¦ by dπ₯ equals six-fifths over negative 40. And we can actually bring the negative on the denominator out to the front of the fraction.

Now, in order to simplify this, one way we could think of this fraction is as negative six-fifths divided by 40 or equivalently negative six-fifths divided by 40 over one. And we can divide these two fractions by changing the divide to a times and flipping the second fraction. This gives us negative six-fifths times one over 40, which is negative six over 200. We can then simplify this by canceling down to give us negative three over 100. And letβs then write this in decimal form. Of course, negative three divided by 100 is negative 0.03. So that gives us our final answer.