- WE WANT TO DETERMINE THE 
DERIVATIVE OF THE GIVEN FUNCTION
USING THE LIMIT DEFINITION 
OF THE DERIVATIVE
AND THEN DETERMINE F PRIME OF 1 
AND F PRIME OF -2,
AND THESE WILL GIVE US THE SLOPE 
OF THE TANGENT LINE
AT X = 1 AND X = -2.
SO HERE IS THE LIMIT DEFINITION 
OF THE DERIVATIVE,
SO TO DETERMINE THE DERIVATIVE 
USING THIS METHOD
WE WILL HAVE TO EVALUATE 
THIS LIMIT.
SO F PRIME OF X
IS GOING TO BE EQUAL 
TO THE LIMIT
AS H APPROACHES 0 
OF THIS DIFFERENCE QUOTIENT
WHERE OUR NUMERATOR IS F OF X 
+ H - F OF X,
AND OUR DENOMINATOR IS H.
SO FOR F OF X + H WE ARE GOING 
TO REPLACE X WITH X + H,
SO WE WILL HAVE 2 ALL OVER X + H
- F OF X WHICH IS JUST 
2 OVER X DIVIDED BY H.
NOW, IN ITS CURRENT FORM 
WE CAN'T DETERMINE THIS LIMIT
BECAUSE IF WE TRY TO PERFORM 
DIRECT SUBSTITUTION
WE HAVE DIVISION BY 0,
SO WE HAVE TO MANIPULATE 
THIS QUOTIENT
AND THEN HOPEFULLY 
WE CAN EVALUATE THIS
USING DIRECT SUBSTITUTION.
SO WHAT WE'RE GOING TO DO
IS MULTIPLY THE NUMERATOR 
AND DENOMINATOR
OF THIS DIFFERENCE QUOTIENT
BY THE COMMON DENOMINATOR 
OF THESE TWO FRACTIONS HERE,
AND THE LEAST COMMON DENOMINATOR 
WOULD BE THE PRODUCT
OF THESE TWO DENOMINATORS,
SO WE ARE GOING TO MULTIPLY THIS 
FRACTION BY X x X + H/X x X + H.
BUT SINCE THESE ARE 
IN FRACTION FORM,
WE'RE GOING TO WRITE THIS 
ALL OVER 1.
NOW LET'S SEE 
WHAT THIS GIVES US.
WE'RE GOING TO LEAVE OUR 
DENOMINATOR IN FACTORED FORM,
SO WE'RE GOING TO HAVE 
HX x X + H,
AND THEN WE NEED TO BE CAREFUL 
WHEN WE MULTIPLY THE NUMERATOR.
WHEN WE FIND THIS PRODUCT HERE,
NOTICE HOW THIS X + H 
WILL SIMPLIFY OUT
WITH THIS X + H HERE, 
AND WE ARE LEFT WITH 2 x X.
SO WE HAVE 2X HERE,
AND THEN WHEN WE DETERMINE 
THIS PRODUCT HERE,
NOTICE HOW THIS X SIMPLIFIES OUT 
WITH THIS X.
SO WE HAVE -2 x (X + H).
NOW LET'S GO AHEAD 
AND CLEAR THESE PARENTHESES
AND COMBINE LIKE TERMS.
SO WE'D HAVE 2X - 2X - 2H 
ALL OVER H x X x X + H.
WELL, 2X - 2X = 0,
SO WE'RE LEFT WITH -2H 
ALL OVER HX x X + H.
NOTICE HOW WE HAVE 
A COMMON FACTOR OF H
BETWEEN THE NUMERATOR 
AND DENOMINATOR
WHICH SIMPLIFIES OUT,
SO NOW WE HAVE THE LIMIT AS H 
APPROACHES 0 OF JUST -2
ALL OVER X x X + H, 
BUT IF H APPROACHES 0,
OUR DENOMINATOR IS GOING 
TO APPROACH
JUST X x X OR X SQUARED.
SO OUR DERIVATIVE IS 
-2 DIVIDED BY X SQUARED.
NOW THAT WE HAVE THE DERIVATIVE
LET'S GO AHEAD AND EVALUATE IT 
AT +1 AND -2 ON THE NEXT PAGE.
SO F PRIME OF 1 IS GOING TO BE 
= -2 ALL OVER 1 SQUARED
WHICH IS -2.
NOW WE WANT F PRIME OF -2,
SO WE'LL HAVE -2 
ALL OVER -2 SQUARED.
-2 OVER +4, 
SO THAT'S EQUAL TO -1/2.
SO THIS TELLS US THE SLOPE OF 
THE TANGENT LINE AT X = 1 = -2,
AND THE SLOPE OF THE TANGENT 
LINE AT -2 = -1/2.
LET'S GO AHEAD 
AND VERIFY THIS GRAPHICALLY.
NOTICE AT X = 1
WE HAVE THIS BLUE TANGENT LINE 
HERE WITH A SLOPE OF -2,
AND NOTICE AT X = -2
WE HAVE THIS BLUE TANGENT LINE 
WITH A SLOPE OF -1/2.
WE WILL TAKE A LOOK AT ONE 
MORE EXAMPLE
INVOLVING A SQUARE ROOT FUNCTION 
IN THE NEXT VIDEO.
I HOPE YOU FOUND THIS HELPFUL.
