# Question Video: Evaluating an Expression Containing Derivatives of Rational Functions Mathematics • Higher Education

Calculate π₯βΆ(dπ¦/dπ₯), given that π¦ = (4π₯β΅ β 5)/(8π₯β΅).

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### 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.

Weβre asked to calculate π₯ raised to the power of six times dπ¦ by dπ₯. And weβre 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, weβre going to need to find the derivative of π¦ with respect to π₯. Thatβs dπ¦ by dπ₯. We could do this by dividing and then using the power rule for differentiation. However, weβre 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₯, thatβs 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 πth power, then π prime of π₯, thatβs 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 letβs clear some space, and letβs 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 thatβs 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, weβre actually asked for π₯ raised to the power of six times dπ¦ by dπ₯. And thatβs 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 weβre left with 200 over 64. In its simplest form, thatβs 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.