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If π¦ is equal to π₯ plus five over π₯ minus five minus π₯ minus five over π₯ plus five, find dπ¦ by dπ₯.
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Our function, π¦, consists of two rational expressions, π₯ plus five over π₯ minus five and π₯ minus five over π₯ plus five.
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And we could find dπ¦ by dπ₯ by using the quotient rule on these two rational expressions.
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However, this would require using the quotient rule twice.
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We can make our work a little easier by combining the two rational expressions into one.
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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.
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We can expand the brackets and then simplify to obtain that π¦ is equal to 20π₯ over π₯ squared minus 25.
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And now, our function consists of only one rational expression.
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Weβre ready to use the quotient rule to differentiate this function.
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The quotient rule tells us that π’ over π£ prime is equal to π£π’ prime minus π’π£ prime over π£ squared.
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Setting π¦ equal to π’ over π£, we obtain that π’ is equal to 20π₯ and π£ is equal to π₯ squared minus 25.
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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₯.
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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.
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We simplify this to obtain that dπ¦ by dπ₯ is equal to negative 20π₯ minus 500 all over π₯ squared minus 25 squared.