### 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. Weβre 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.