- 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.