### Video Transcript

Calculate π₯ raised to the power of six times dπ¦ by dπ₯ given that π¦ is equal to four times π₯ raised to the power of five minus five all over eight times π₯ raised to the power of five.

Weβre asked to calculate π₯ raised to the power of six times dπ¦ by dπ₯. And weβre given that π¦ is a rational function four π₯ raised to the power of five minus five over eight times π₯ raised to the power of five. In order to calculate this, weβre going to need to find the derivative of π¦ with respect to π₯. Thatβs dπ¦ by dπ₯. We could do this by dividing and then using the power rule for differentiation. However, weβre going to use the quotient rule to show how that works. And this says that if π of π₯ is the ratio of π’ of π₯ over π£ of π₯, then the derivative of π with respect to π₯, thatβs π prime of π₯, is equal to π’ prime of π₯ times π£ of π₯ minus π’ of π₯ times π£ prime of π₯ all over π£ squared.

In our case, π’ of π₯ is equal to four π₯ raised to the power five minus five. And π£ of π₯ is equal to eight times π₯ raised to the power of five. And to find dπ¦ by dπ₯, we need to find π’ prime of π₯ and π£ prime of π₯. So with π’ of π₯ equal to four π₯ raised to the power of five minus five, we can find dπ’ by dπ₯ by using the power rule for differentiation. This says that if π of π₯ is a function of the form π times π₯ raised to the πth power, then π prime of π₯, thatβs dπ by dπ₯, is equal to π times π times π₯ raised to the power of π minus one. That is, we multiply by the exponent π and then subtract one from the exponent.

In our first term, our exponent is five so that the derivative is five times four times π₯ raised to the power of four. And we know that the derivative of the constant negative five is zero so that π’ prime of π₯ is 20 times π₯ raised to the power of four. Similarly, for π£, our exponent is five again so that π£ prime of π₯ is five times eight times π₯ raised to the power of four, that is, 40 times π₯ raised to the power of four.

So now letβs clear some space, and letβs use π’, π£, and their derivatives in the quotient rule to find dπ¦ by dπ₯. In our numerator, we have π’ prime times π£ minus π’ times π£ prime. And in our denominator, we have eight times π₯ raised to the power of five all squared, which is π£ squared. In our numerator, we have a common factor of 20π₯ raised to the power of four. And we can take this outside of our bracket to get 20π₯ raised to the power of four times eight π₯ raised to the power of five minus two times four π₯ raised to the power of five minus five all over eight π₯ to the power of five squared.

And now, if we multiply out our internal bracket, in the numerator, we can see that the positive and negative eight π₯ to the power of five cancel each other out. So we have 10 times 20π₯ raised to the power of four over eight times π₯ raised to the power of five all squared. And thatβs 200π₯ raised to the power of four times eight π₯ raised to the power of five all squared. And now, if we multiply out our denominator, we have dπ¦ by dπ₯ is equal to 200π₯ raised to the power of four over 64 times π₯ raised to the 10th power.

Now, if we refer back to our question, weβre actually asked for π₯ raised to the power of six times dπ¦ by dπ₯. And thatβs equal to π₯ raised to the power of six times 200 times π₯ raised to the power of four over 64 times π₯ raised to the 10th power. And recalling that π₯ raised to the power of π times π₯ raised to the power of π is π₯ raised to the power of π plus π, where in our case π is six and π is four, we have 200 times π₯ raised to the 10th power over 64 times π₯ raised to the 10th power. And so we can cancel our π₯ to the 10th powers and weβre left with 200 over 64. In its simplest form, thatβs 25 over eight.

And so, given π¦ equal to four times π₯ raised to the power of five minus five divided by eight times π₯ to the power of five, then π₯ raised to the power of six times dπ¦ by dπ₯ is equal to 25 over eight.