- WE WANT TO DETERMINE THE 
DERIVATIVE OF THE GIVEN FUNCTION
USING THE LIMIT DEFINITION 
OF THE DERIVATIVE
AND THEN DETERMINE THE SLOPE 
OF THE TANGENT LINE AT X = -1.
SO HERE IS OUR LIMIT DEFINITION 
OF THE DERIVATIVE.
NOTICE HERE WE ARE USING H.
SOMETIMES YOU'LL SEE SOME 
TEXTBOOKS USING DELTA X,
BUT I THINK WHEN YOU ARE FIRST 
LEARNING HOW TO DO THIS
IT IS EASIER TO KEEP 
X's AND H's APART
FROM X's AND DELTA X's.
HERE WE'RE GIVEN 
Y = X SQUARED - X
WHICH IS THE SAME AS SAYING 
F OF X = X SQUARED - X.
SO THE DERIVATIVE 
OF OUR FUNCTION
IS GOING TO BE EQUAL 
TO THE LIMIT
AS H APPROACHES 0 
OF THIS DIFFERENCE QUOTIENT.
OUR NUMERATOR IS GOING TO BE 
F OF X + H,
WHICH MEANS WE'RE GOING TO TAKE 
OUR FUNCTION
AND REPLACE EACH X WITH X + H,
SO WE WILL HAVE X + H SQUARED 
- X + H,
AND THIS IS JUST F OF X + H.
AND NOW WE HAVE TO SUBTRACT 
F OF X.
WE HAVE TO BE CAREFUL HERE
BECAUSE WE HAVE TO SUBTRACT 
THE ENTIRE FUNCTION,
SO I'LL HAVE 
THE QUANTITY X SQUARED - X.
IF WE LEAVE OFF 
THESE PARENTHESES,
IT WOULD NOT BE CORRECT.
WE NEED TO DIVIDE 
ALL OF THIS BY H.
AND JUST TO DOUBLE-CHECK 
BEFORE WE DO ALL THIS WORK,
NOTICE THE NUMERATOR IS F OF X 
+ H - F OF X.
SO ALL OF THIS IS F OF X + H 
AND THIS IS F OF X.
NOW THAT WE HAVE VERIFIED 
IT'S SET UP CORRECTLY,
WE CAN START TO SIMPLIFY 
THIS QUOTIENT.
NOW WE HAVE TO MULTIPLY 
ALL THIS OUT,
COMBINE LIKE TERMS, 
AND TRY TO SIMPLIFY OUT THIS H.
RIGHT NOW THE H IS A PROBLEM
BECAUSE IF WE TRY TO PERFORM 
DIRECT SUBSTITUTION
OUR DENOMINATOR WOULD BE 0.
NOW I KNOW IT'S EASY TO FORGET,
BUT IT IS IMPORTANT THAT WE KEEP 
WRITING THIS LIMIT NOTATION.
THE LIMIT AS H APPROACHES 0.
NOW IF WE MULTIPLY 
(X + H) x (X + H)
WE ARE GOING TO HAVE X SQUARED 
+ (2HX + H SQUARED).
NOW JUST IN CASE I LOST 
YOU THERE, IT LOOKS LIKE THIS.
WE HAVE X SQUARED + (HX + HX), 
THAT'S 2HX,
AND THEN + H SQUARED.
HERE WE ARE GOING TO CLEAR 
THE PARENTHESES
WHICH YOU CAN THINK OF AS 
DISTRIBUTING A -1,
SO WE WILL HAVE -X - H, AND 
THEN WE WILL DO THE SAME HERE.
WE'RE GOING TO HAVE - X SQUARED 
+ X DIVIDED BY H.
NOW GOOD THINGS 
START TO HAPPEN HERE.
NOTICE HOW WE HAVE AN X SQUARED 
- X SQUARED, THAT'S 0,
AND WE ALSO HAVE -X + X.
AND NOW IF WE LOOK AT ALL 
OF THE TERMS IN THE NUMERATOR,
NOTICE HOW EACH OF THEM 
HAVE A COMMON FACTOR OF H,
SO NOW WE'LL FACTOR OUT AN H 
FROM THE NUMERATOR.
SO WE HAVE THE LIMIT AS H 
APPROACHES 0, WE'LL HAVE H.
SO IF WE FACTOR OUT AN H 
FROM 2HX WE HAVE 2X.
FACTOR OUT 1 FACTOR OF H 
FROM HERE, WE HAVE + H.
IF WE FACTOR OUT H FROM H 
WE HAVE 1/H.
AND NOW NOTICE THAT WE HAVE 
A COMMON FACTOR OF H
BETWEEN THE NUMERATOR 
AND DENOMINATOR
WHICH SIMPLIFIES OUT, 
LEAVING US WITH THE LIMIT
AS H APPROACHES 0 
OF JUST 2X + H - 1.
WELL, AS H APPROACHES 0 THIS 
IS THE ONLY TERM AFFECTED BY H,
SO AS H APPROACHES 0, 
THIS LIMIT = 2X - 1
WHICH IS F PRIME OF X 
OR A DERIVATIVE OF F OF X.
SO NOW LET'S GO AHEAD
AND TAKE THIS INFORMATION 
ON THE NEXT SLIDE
TO DETERMINE THE SLOPE 
OF THE TANGENT LINE.
SO IF F PRIME OF X = 2X - 1,
THE SLOPE OF THE TANGENT LINE 
AT X = -1 WILL BE F PRIME OF -1,
SO WE HAVE 2 x -1 - 1, 
SO F PRIME OF -1 = -3.
SO THE SLOPE OF THE TANGENT LINE
AT X = -1 IS -3.
LET'S GO AHEAD AND VERIFY 
THAT GRAPHICALLY.
IN RED WE HAVE THE GRAPH 
OF THE GIVEN FUNCTION,
AND SO WHAT WE FOUND WAS 
THE SLOPE OF THIS TANGENT LINE
WHICH IS EQUAL TO -3.
AND THERE'S ONE MORE THING 
I DO WANT TO MENTION.
IF YOU HAD TO DETERMINE 
THE EQUATION
OF THE TANGENT LINE AT X = 1,
NOTICE HOW YOU WOULD HAVE TO 
DETERMINE THE POINT OF TANGENCY.
SO TO DETERMINE THE POINT 
OF TANGENCY,
WE WOULD HAVE TO SUBSTITUTE X 
= -1 INTO THE ORIGINAL FUNCTION
TO DETERMINE THE Y COORDINATE 
ON THE RED GRAPH.
SO F OF -1 = 2 WHICH TELLS US 
THE POINT OF TANGENCY IS (-1,2).
IT IS VERY IMPORTANT THAT 
WE KEEP THIS STRAIGHT.
SUBBING X = -1 INTO THE FUNCTION
GIVES US THE Y COORDINATE 
OF THE POINT OF TANGENCY,
BUT SUBBING -1 INTO OUR 
DERIVATIVE FUNCTION
GIVES US THE SLOPE 
OF THE TANGENT LINE.
WE WILL LOOK AT MORE EXAMPLES 
IN THE NEXT SEVERAL VIDEOS.
