- IF F OF X EQUALS THE QUANTITY 3X + 1 DIVIDED BY THE QUANTITY
X - 2 WE WANT TO FIND F PRIME OF X, THE DERIVATIVE FUNCTION
AND F PRIME OF 1.
BECAUSE OUR FUNCTION IS A QUOTIENT,
WE'LL HAVE TO APPLY THE QUOTIENT RULE IN ORDER
TO FIND OUR DERIVATIVE FUNCTION.
THE QUOTIENT RULE IS STATED HERE
WHERE IF WE WANT TO FIND
THE DERIVATIVE OF A QUOTIENT FUNCTION WHERE THE NUMERATOR
IS FUNCTION F AND THE DENOMINATOR IS FUNCTION G
THE DERIVATIVE IS EQUAL TO G x F PRIME - F x G PRIME
DIVIDED BY G SQUARED.
SO WE HAVE THE DENOMINATOR
x THE DERIVATIVE OF THE NUMERATOR - THE NUMERATOR
x THE DERIVATIVE OF THE DENOMINATOR
ALL DIVIDED BY THE DENOMINATOR SQUARED.
SO AS YOU CAN SEE THE QUOTIENT RULE IS FAIRLY INVOLVED.
ONE WAY TO HELP REMEMBER THE QUOTIENT RULE
IS GIVEN HERE IN BLUE WHERE IF BIG D REPRESENTS THE DERIVATIVE,
HIGH REPRESENTS THE FUNCTION IN THE NUMERATOR
AND LOW REPRESENTS THE FUNCTION IN THE DENOMINATOR
WE CAN STATE THE QUOTIENT RULE AS LOW D HIGH - HIGH D
LOW LOW LOW.
ONCE AGAIN, WE HAVE LOW D HIGH - HIGH D LOW LOW LOW.
SO IF WERE TO SAY THIS OR EVEN SING IT ABOUT 10 TIMES
YOU WOULD PROBABLY REMEMBER THE QUOTIENT RULE
WHERE AGAIN WE HAVE THE DENOMINATOR
OR THE LOW FUNCTION x THE DERIVATIVE OF THE HIGH FUNCTION
OR THE NUMERATOR - THE HIGH FUNCTION OR THE NUMERATOR
x THE DERIVATIVE OF THE LOW FUNCTION
ALL DIVIDED BY LOW LOW WHICH IS THE DENOMINATOR SQUARED.
SO LOOKING AT OUR GIVEN FUNCTION IF F OF X = THE QUANTITY 3X + 1
DIVIDED BY THE QUANTITY X - 2 USING THE FORMAL QUOTIENT RULE
WE'D LET THE NUMERATOR BE EQUAL TO F
AND THE DENOMINATOR BE EQUAL TO G
OR IF WE WANT TO USE HIGH AND LOW INSTEAD THE NUMERATOR
WOULD BE HIGH AND THE DENOMINATOR WOULD BE LOW.
SO WHICH EVER FORMULA WE WANT TO USE F PRIME OF X
IS GOING TO BE EQUAL TO THE FUNCTION WHERE THE DENOMINATOR
IS JUST THE DENOMINATOR SQUARED.
SO WE'D HAVE THE QUANTITY X - 2 SQUARED.
THEN THE NUMERATOR IS GOING TO BE THE DENOMINATOR
x THE DERIVATIVE OF THE NUMERATOR - THE NUMERATOR
x THE DERIVATIVE OF THE DENOMINATOR.
SO AGAIN WE CAN LOOK AT THIS AS G x F PRIME - F x G
PRIME DIVIDED BY G SQUARED OR LOW D HIGH - HIGH D LOW LOW LOW.
SO NOTICE HOW FOR THIS FIRST STEP
WE WROTE OUT THE QUOTIENT RULE BUT DIDN'T FIND ANY DERIVATIVES.
SO NOW WE'LL GO BACK AND FIND THE DERIVATIVE HERE,
HERE AND BEGIN TO SIMPLIFY.
SO F PRIME OF X IS EQUAL TO--
THE DENOMINATOR IS GOING TO STAY THE SAME
AND THEN WE'LL HAVE THE QUANTITY
X - 2 x THE DERIVATIVE OF 3X + 1.
THE DERIVATIVE OF 3X WOULD JUST BE 3,
THE DERIVATIVE OF A 1 IS 0 - THE QUANTITY 3X + 1
x THE DERIVATIVE OF X - 2.
THE DERIVATIVE OF X IS 1, THE DERIVATIVE OF -2 IS 0.
SO WE'D JUST HAVE 1 AND NOW WE'LL GO AND DISTRIBUTE
AND COMBINE LIKE TERMS.
SO WE HAVE F PRIME OF X IS EQUAL TO--
AGAIN THE DENOMINATOR STAYS THE SAME.
IF WE DISTRIBUTE HERE WE WOULD HAVE 3X - 6.
WE NEED TO BE CAREFUL HERE
BECAUSE WE'RE SUBTRACTING THIS PRODUCT.
HERE WE'RE JUST DISTRIBUTING 1
WHICH ISN'T GOING TO CHANGE ANYTHING.
WE HAVE - THE QUANTITY 3X + 1 WHICH WOULD BE - 3X - 1
AND NOTICE HOW THE X TERMS ARE OPPOSITES. 3X - 3X IS 0.
SO OUR DERIVATIVE FUNCTION,
F PRIME OF X IS GOING TO BE -7
DIVIDED BY THE QUANTITY X - 2 SQUARED.
NOW THE SECOND PART OF THIS QUESTION ASKS US TO FIND
F PRIME OF 1 SO NOW THAT WE HAVE OUR DERIVATIVE FUNCTION
WE'LL SUBSTITUTE 1 FOR X
AND EVALUATE THE DERIVATIVE FUNCTION.
REMEMBER THIS WILL TELL US THE SLOPE OF THE TANGENT LINE
WHEN X = 1.
LET'S GO AHEAD AND DO THIS ON THE NEXT SLIDE.
SO F PRIME OF 1 IS GOING TO BE EQUAL TO -7
DIVIDED BY THE QUANTITY 1 - 2 SQUARED, THIS IS -7.
THIS WILL BE -1 SQUARED.
SO F PRIME OF 1 IS EQUAL TO -7.
LET'S GO AHEAD AND VERIFY THIS BY LOOKING
AT THE GRAPH OF OUR FUNCTION.
SO HERE'S THE GRAPH OF OUR FUNCTION.
NOTICE WHEN X = 1 WE'D BE AT THIS POINT HERE ON THE FUNCTION
AND SINCE WE FOUND F PRIME OF 1 = -7
IF WE SKETCH A TANGENT LINE AT THIS POINT
IT WOULD LOOK SOMETHING LIKE THIS.
IT'S A LITTLE HARD TO TELL
BUT IT DOES LOOK LIKE THE SLOPE OF THIS TANGENT LINE
WOULD BE EQUAL TO -7.
NOTICE HOW THE Y AXIS IS SCALED BY FOURS.
OKAY, I HOPE YOU FOUND THIS INTRODUCTORY VIDEO
ON THE QUOTIENT RULE HELPFUL.