- WE WANT TO DETERMINE THE DERIVATIVE
OF THE GIVEN FUNCTION,
WHICH IS A QUOTIENT OF TWO DIFFERENTIABLE FUNCTIONS.
SO IF WE LET THE NUMERATOR EQUAL F AND THE DENOMINATOR EQUAL G
WE CAN DETERMINE THIS DERIVATIVE
BY APPLYING THE QUOTIENT RULE OF DIFFERENTIATION GIVEN HERE.
SO IF WE'RE TRYING TO DIFFERENTIATE A FUNCTION
WHERE THE NUMERATOR IS GIVEN BY F
AND THE DENOMINATOR IS GIVEN BY G
THE DERIVATIVE IS EQUAL TO G x F PRIME - F x G PRIME
ALL OVER G SQUARED.
IT DOES TAKE SOME TIME TO MEMORIZE THIS FORMULA
SO HERE'S A NICE WAY TO REMEMBER IT.
IF WE THINK OF THE NUMERATOR AS THE HIGH FUNCTION
AND THE DENOMINATOR AS THE LOW FUNCTION
AND D HIGH AS THE DERIVATIVE OF THE NUMERATOR
AND D LOW AS THE DERIVATIVE OF THE DENOMINATOR
WE COULD STATE THE QUOTIENT RULE AS LOW D HIGH - HIGH D LOW
LOW LOW WHICH MEANS THE EXACT SAME THING AS THE FORMULA HERE
GIVEN IN RED.
NOW BEFORE WE APPLY THE QUOTIENT RULE
I DO WANT TO MENTION THAT SINCE WE'RE DIVIDING BY A MONOMIAL
WE COULD DIVIDE EACH TERM IN THE NUMERATOR BY X SQUARED
AND THEN JUST APPLY THE POWER RULE
WHICH WE WILL DO ON THE NEXT SLIDE
BUT USING A FUNCTION LIKE THIS IS A NICE WAY TO GET USED TO
USING THE QUOTIENT RULE.
SO F PRIME OF X IS GOING TO BE EQUAL TO--
I ALWAYS LIKE TO START WITH THE DENOMINATOR
BECAUSE THE DENOMINATOR
IS JUST THE ORIGINAL DENOMINATOR SQUARED.
SO WE'LL HAVE X TO THE 2nd SQUARED IN OUR DENOMINATOR,
AND AGAIN THE NUMERATOR IS GOING OT BE G x F PRIME
OR THE DENOMINATOR x THE DERIVATIVE OF THE NUMERATOR
AND THE DERIVATIVE OF X TO THE 5th + 2
WOULD BE 5X TO THE 4th
MINUS F x G PRIME OR THE NUMERATOR
x THE DERIVATIVE OF THE DENOMINATOR WHICH WOULD BE 2X.
SO NOW WE'RE DONE WITH THE CALCULUS.
WE FOUND THE DERIVATIVE
SO NOW WE WANT TO SIMPLIFY THIS FUNCTION.
SO WE'LL START BY ELIMINATING THE PARENTHESES.
THE DENOMINATOR IS GOING TO BE X TO THE 4th.
HERE WE'LL HAVE 5 X TO THE 6th.
NOW WE DO HAVE TO BE CAREFUL HERE
BECAUSE WE'RE SUBTRACTING THIS PRODUCT.
SO WE'LL DISTRIBUTE AND THEN SUBTRACT THE TERM.
SO WE HAVE 2X x X TO THE 5th THAT'S 2X TO THE 6th
BUT WE'RE SUBTRACTING THAT SO WE'LL HAVE - 2X TO THE 6th
AND THEN 2X x 2 IS 4X BUT IT'S - 4X.
NOTICE HOW WE HAVE TWO LIKE TERMS HERE
SO LET'S COMBINE THOSE.
WE'LL HAVE 3X TO THE 6th - 4X
AND NOW THAT WE'RE DIVIDING BY A MONOMIAL
WE'LL DIVIDE EACH TERM IN THE NUMERATOR BY X TO THE 4th.
SO F PRIME OF X IS EQUAL TO 3X TO THE 6th
DIVIDE BY X TO THE 4th - 4X DIVIDED BY X TO THE 4th.
SO HERE FOUR FACTORS OF X WILL SIMPLIFY OUT.
SO WE HAVE 3X SQUARED
AND HERE ONLY ONE FACTOR OF X WILL SIMPLIFY OUT
LEAVING THREE FACTORS OF X IN THE DENOMINATOR.
SO WE HAVE - 4 DIVIDED BY X TO THE 3rd.
SO HERE'S OUR DERIVATIVE FUNCTION
USING THE QUOTIENT RULE.
NOW I DO WANT TO SHOW THIS SAME PROBLEM
WITHOUT USING THE QUOTIENT RULE
JUST TO SHOW THE RESULT IS THE SAME.
SO NOW INSTEAD OF USING THE QUOTIENT RULE
WE'LL DIVIDE EACH TERM IN THE NUMERATOR BY X TO THE 2nd.
NOW THIS ONLY WORKS WHEN WE'RE DIVIDING BY A MONOMIAL.
SO WE'D HAVE X TO THE 3rd
AND HERE WE'D HAVE + 2 X TO THE POWER OF -2
SO THAT WE CAN APPLY THE POWER RULE.
SO NOW OUR DERIVATIVE FUNCTION WOULD BE 3X SQUARED.
THIS WOULD BE - 4 X TO THE-- -2 - 1 IS -3
AND THE LAST STEP WOULD BE TO SIMPLIFY THIS LAST TERM.
SO WE HAVE 3X SQUARED.
THIS WOULD BE - 4 DIVIDED BY X TO THE 3rd.
SO AS WE EXPECT THE DERIVATIVE FUNCTION IS THE SAME
REGARDLESS OF THE TECHNIQUE THAT WE USE.
NOW IT DOES SEEM LIKE THIS METHOD WAS A LOT EASIER
FOR THIS PROBLEM
BUT THE IDEA ON THIS WAS TO INTRODUCE THE QUOTIENT RULE
BECAUSE THERE ARE MANY QUOTIENTS
WHERE WE DO HAVE TO APPLY THE QUOTIENT RULE.
I HOPE THIS WAS HELPFUL.