- 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.