# Video: Differentiating a Combination of Rational Functions Using the Quotient Rule

If π¦ = ((π₯ + 5)/(π₯ β 5)) β ((π₯ β 5)/(π₯ + 5)), find dπ¦/dπ₯.

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### Video Transcript

If π¦ is equal to π₯ plus five over π₯ minus five minus π₯ minus five over π₯ plus five, find dπ¦ by dπ₯.

Our function, π¦, consists of two rational expressions, π₯ plus five over π₯ minus five and π₯ minus five over π₯ plus five. And we could find dπ¦ by dπ₯ by using the quotient rule on these two rational expressions. However, this would require using the quotient rule twice. We can make our work a little easier by combining the two rational expressions into one. We obtain that π¦ is equal to π₯ plus five squared minus π₯ minus five squared all over π₯ minus five times π₯ plus five. We can expand the brackets and then simplify to obtain that π¦ is equal to 20π₯ over π₯ squared minus 25.

And now, our function consists of only one rational expression. Weβre ready to use the quotient rule to differentiate this function. The quotient rule tells us that π’ over π£ prime is equal to π£π’ prime minus π’π£ prime over π£ squared. Setting π¦ equal to π’ over π£, we obtain that π’ is equal to 20π₯ and π£ is equal to π₯ squared minus 25. Next, we can find π’ prime and π£ prime, which gives us that π’ prime is equal to 20 and π£ prime is equal to two π₯.

Now, we can substitute them into the quotient rule in order to find that dπ¦ by dπ₯ is equal to π₯ squared minus 25 times 20 minus 20π₯ times two π₯ all over π₯ squared minus 25 squared. We simplify this to obtain that dπ¦ by dπ₯ is equal to negative 20π₯ minus 500 all over π₯ squared minus 25 squared.