- GIVEN F OF X = G OF X DIVIDED 
BY H OF X
WE WANT TO USE THE GRAPHS 
PROVIDED
TO DETERMINE F PRIME OF 5.
WELL NOTICE FIRST THAT FUNCTION 
F IS A QUOTIENT OF TWO FUNCTIONS
SO WE'LL FIRST FIND
THE DERIVATIVE FUNCTION
USING THE QUOTIENT RULE 
PROVIDED HERE BELOW
AND THEN WE'LL COME BACK 
AND ANALYZE THE GRAPH
TO DETERMINE F PRIME OF 5.
SO WE'LL COME BACK TO THESE 
GRAPHS IN JUST A MOMENT.
REVIEWING THE QUOTIENT RULE
IF WE WANT THE DERIVATIVE 
OF FUNCTION F
DIVIDED BY FUNCTION G 
WITH RESPECT TO X
THIS DERIVATIVE WOULD BE EQUAL 
TO G x F PRIME - F x G PRIME
DIVIDED BY G SQUARED.
AGAIN, WHERE F 
IS IN THE NUMERATOR
AND G IS IN THE DENOMINATOR.
SO WE HAVE TO BE A LITTLE 
CAREFUL HERE
BECAUSE OUR VARIABLES 
ARE DIFFERENT.
OUR NUMERATOR IS G OF X 
AND OUR DENOMINATOR IS H OF X.
SO IT MIGHT BE MORE HELPFUL TO 
THINK OF THIS AS THE DENOMINATOR
x THE DERIVATIVE 
OF THE NUMERATOR - THE NUMERATOR
x THE DERIVATIVE 
OF THE DENOMINATOR
DIVIDED BY THE DENOMINATOR 
SQUARED.
SO IN OUR CASE, F PRIME OF X IS 
GOING TO BE EQUAL TO A QUOTIENT
BUT NOTICE HOW THE DENOMINATOR 
IS JUST THE DENOMINATOR SQUARED.
SO WE WOULD HAVE H OF X SQUARED
AND THE NUMERATOR 
IS THE DENOMINATOR
x THE DERIVATIVE OF THE 
NUMERATOR WHICH WOULD BE H OF X
x G PRIME OF X
AND THEN - THE NUMERATOR x THE 
DERIVATIVE OF THE DENOMINATOR.
SO WE'D HAVE G OF X 
x H PRIME OF X.
BUT OUR GOAL IS TO FIND 
F PRIME OF 5
SO NOW WE'LL SUBSTITUTE 5 FOR X.
SO WE'LL HAVE H OF 5 x G PRIME 
OF 5 - G OF 5 x H PRIME OF 5
DIVIDED BY H OF 5 SQUARED.
LET'S GO AHEAD AND TAKE THIS 
DERIVATIVE FUNCTION VALUE
TO THE PREVIOUS SLIDE
AND ANALYZE THE GRAPHS.
AGAIN, HERE'S THE GRAPH OF 
FUNCTION G OF X
AND HERE'S THE GRAPH OF FUNCTION 
H OF X.
NOTICE HOW WE NEED TO FIND G OF 
5 AND G PRIME OF 5
WHERE G PRIME OF 5 WOULD BE THE 
SLOPE OF TANGENT LINE AT X = 5.
WE ALSO HAVE TO FIND H OF 5 
AND H PRIME OF 5
WHERE H PRIME OF 5 WOULD BE THE 
SLOPE OF THE TANGENT LINE
AT X = 5 FOR H OF X.
LET'S BEGIN BY DETERMINING 
THE FUNCTION VALUES G OF 5
AND H OF 5
AND THEN WE'LL TALK ABOUT 
HOW WE'RE GOING TO FIND
THESE DERIVATIVE FUNCTION 
VALUES.
SO TO EVALUATE G OF 5
WE WANT TO FIND THE FUNCTION 
VALUE OR Y VALUE WHEN X = 5.
HERE'S WHERE X = 5.
NOTICE HOW THE FUNCTION VALUE 
WOULD BE 4
AND THEN FOR H OF 5 
HERE'S WHERE X = 5.
NOTICE HOW THE FUNCTION VALUE 
IS 3
AND NOW LET'S DISCUSS HOW WE'RE 
GOING TO FIND
THESE DERIVATIVE FUNCTION 
VALUES.
AGAIN, G PRIME OF 5 IS EQUAL TO 
THE SLOPE OF THE TANGENT LINE
AT X EQUALS 5.
SO HERE'S THE POINT WHERE X = 5
AND NOTICE HOW IF WE SKETCH 
A TANGENT LINE TO OUR FUNCTION
BECAUSE IT'S LINEAR
THE TANGENT LINE WILL CONTAIN 
THE GRAPH OF THE FUNCTION.
THE GRAPH OF THE TANGENT LINE 
WOULD LOOK LIKE THIS.
SO BECAUSE OUR FUNCTION 
IS LINEAR
AND THE TANGENT LINE CONTAINS 
THE FUNCTION
WE CAN USE THE GRAPH 
TO DETERMINE THE SLOPE
OF THIS TANGENT LINE
AND THEREFORE FIND G PRIME OF 5.
SO IF WE SELECT THIS POINT HERE 
THAT'S ALSO ON THE TANGENT LINE
NOTICE HOW THE CHANGE OF Y IS +1 
BECAUSE WE GO UP 1,
AND THE CHANGE OF X IS +2 
BECAUSE WE GO RIGHT 2
TO GO FROM THIS POINT 
TO THIS POINT
AND SINCE THE SLOPE OF THE 
TANGENT LINE IS 1/2 AT X = 5
G PRIME OF 5 IS 1/2.
NOW TAKING A LOOK AT FUNCTION H,
IF WE SKETCH A TANGENT LINE 
AT X = 5 IT WOULD LOOK LIKE THIS
CONTAINING THE RIGHT PIECE 
OF THE FUNCTION
AND THEREFORE ONCE AGAIN
WE CAN SELECT TWO POINTS 
ON THIS TANGENT LINE
TO DETERMINE THE SLOPE AND 
THEREFORE FIND H PRIME OF 5.
SO IF WE USE THIS POINT AT X = 5 
AND THIS POINT THAT X = 4
NOTICE HOW WE'D HAVE TO GO DOWN 
ONE UNIT AND RIGHT ONE UNIT.
SO THE CHANGE OF Y IS -1 
AND THE CHANGE OF X IS +1
AND THEREFORE THE SLOPE 
OF THE TANGENT LINE IS -1, X = 5
AND THEREFORE H PRIME OF 5 
IS -1,
AND NOW WE JUST NEED TO PERFORM 
SUBSTITUTION
INTO OUR FUNCTION VALUE HERE
WHICH SHOULD BE NOT JUST F PRIME 
OF X BUT F PRIME OF 5.
SO LET'S GO AHEAD AND DO THAT.
THE DENOMINATOR WOULD BE H OF 5 
SQUARED WHERE H OF 5 IS 3.
SO WE'D HAVE 3 SQUARED.
LOOKING AT THE NUMERATOR WE HAVE 
H OF 5 WHICH WE KNOW IS 3
x G PRIME OF 5 WHICH IS 1/2
AND THEN - G OF 5 WHICH IS 4 
x H PRIME OF 5 WHICH IS -1.
SO WE HAVE 3/2 + 4.
SO I'M GOING TO WRITE + 4/1
AND THEN WE'RE GOING TO DIVIDE 
THIS BY 9.
IF WE OBTAIN A COMMON 
DENOMINATOR HERE,
THAT WOULD BE 2 
SO WE MULTIPLY BY 2/2.
SO THIS IS EQUAL TO 3/2 + 8/2 
THAT WOULD BE 11/2
AND THEN INSTEAD OF DIVIDING 
BY 9
WE CAN MULTIPLY 
BY THE RECIPROCAL.
THE RECIPROCAL OF 9 IS 1/9.
SO WE HAVE x 1/9 AND THEREFORE 
H PRIME OF 5 IS EQUAL TO 11/18.
I HOPE YOU FOUND THIS HELPFUL.
