- WELCOME TO FINDING DERIVATIVES USING THE LIMIT DEFINITION.
NOW, FINDING DERIVATIVES USING THE LIMIT DEFINITION
IS NOT THE MOST POPULAR WAY OF FINDING DERIVATIVES
BUT WE WILL BE USING DERIVATIVE FORMULAS
THAT ARE MUCH EASIER VERY SOON
BUT THIS IS SOMETHING THAT WE HAVE TO EXPERIENCE
TO FULLY APPRECIATE THE SHORTCUTS OF THE FORMULAS.
THE GOAL OF THIS VIDEO
WILL BE TO FIND THE DERIVATIVE OF A FUNCTION
USING THE LIMIT DEFINITION,
FIND THE VALUE OF A DERIVATIVE AT A GIVEN POINT
AND FIND THE EQUATION OF A TANGENT LINE AT A GIVEN POINT.
GIVEN F OF X = X SQUARED + 1
FIND THE DERIVATIVE OF F OF X USING A LIMIT DEFINITION
THEN FIND THE VALUE OF THE DERIVATIVE AT THE POINT (-1,2).
SO WE FIRST NEED TO SETUP OUR LIMIT.
SO WE'LL HAVE THE LIMIT AS DELTA X APPROACHES 0.
NOW F OF X + DELTA X WILL END UP BEING
X + DELTA X SQUARED + 1 - F OF X WHICH WILL BE X SQUARED + 1.
THESE PARENTHESIS ARE IMPORTANT ALL DIVIDED BY DELTA X.
OKAY, SO LET'S EXPAND THIS AND SIMPLIFY.
X + DELTA X SQUARED WOULD GIVE US X SQUARED
+ 2X DELTA X + DELTA X SQUARED.
AND THEN WE HAVE + 1 - X SQUARED - 1 DIVIDED BY DELTA X.
NOW THIS LOOKS MESSY BUT A LOT OF IT WILL SIMPLIFY.
NOTICE WE HAVE X SQUARED HERE - X SQUARED.
HERE WE HAVE + 1 AND THEN WE HAVE -1,
SO THIS SIMPLIFIES INTO
2X DELTA X + DELTA X SQUARED DIVIDED BY DELTA X.
NOW AGAIN WE CAN'T JUST DO DIRECT SUBSTITUTION
OTHERWISE WE'D HAVE AN UNDEFINED VALUE.
WE'D HAVE 0 IN THE DENOMINATOR.
HOWEVER, THERE IS A COMMON FACTOR OF DELTA X
IN THE NUMERATOR WHICH WE CAN FACTOR OUT.
IF WE FACTOR OUT THE DELTA X
YOU CAN SEE THE NUMERATOR AND DENOMINATOR
HAVE A COMMON FACTOR OF DELTA X.
SO OUR FINAL LIMIT WOULD JUST BE THE LIMIT
AS DELTA X APPROACHES 0 OF 2X + DELTA X.
DELTA X IS APPROACHING 0
THEREFORE OUR LIMIT WOULD JUST BE = TO 2X.
SO WHAT WE HAVE FOUND IS A DERIVATIVE OF F OF X IS = TO 2X.
SO IF WE WANNA KNOW THE VALUE
OF THE DERIVATIVE AT THE POINT (-1,2),
WE NEED TO EVALUATE THE DERIVATIVE
AT THE X VALUE OF -1 WHICH = -2.
THIS WOULD BE THE SLOPE OF THE TANGENT LINE
AT THE POINT (-1,2).
LET'S VERIFY THIS WITH A GRAPH.
HERE'S OUR FUNCTION AND HERE'S OUR TANGENT LINE
AND WHAT WE'RE SAYING FROM OUR WORK
IS THAT THE SLOPE OF THIS LINE = -2
AND YOU CAN SEE IF I START HERE
AT THE POINT OF TANGENCY, (-1,2),
I GO DOWN TO RIGHT 1 OF COURSE, THAT POINT IS ALSO ON THE LINE.
THAT VERIFIES OUR WORK IS CORRECT.
ONE OTHER THING WE MIGHT WANNA DO AT THIS POINT
IS TAKE A LOOK AT THE GRAPH OF THE DERIVATIVE FUNCTION AS WELL.
HERE'S OUR FUNCTION F OF X. AGAIN HERE'S OUR TANGENT LINE.
THE GREEN GRAPH IS THE GRAPH OF OUR DERIVATIVE.
THE VALUES OF THE DERIVATIVE WOULD GIVE US
THESE SLOPES OF THE TANGENT LINES.
SO FOR EXAMPLE AT THIS POINT WHERE X = -1
WE SAID OUR DERIVATIVE = -2,
THE SLOPE OF THE TANGENT LINE, DOWN 2, RIGHT 1.
NOTICE ON THE GREEN GRAPH THE Y VALUE = -2,
ANOTHER WAY TO VERIFY OUR WORK.
LET'S TAKE A LOOK AT ANOTHER.
GIVEN F OF X = THE SQUARE ROOT OF X,
FIND F PRIME OF X USING THE LIMIT DEFINITION,
FIND THE VALUE DERIVATIVE AT THE POINT (1,1).
FIRST WE NEED TO SETUP OUR LIMIT,
SO WE HAD THE LIMIT AS DELTA X APPROACHES 0
NOW F OF X + DELTA X
WILL BE THE SQUARE ROOT OF X + DELTA X - -F OF X
WILL BE THE SQUARE ROOT OF X DIVIDED BY DELTA X.
ON THIS LIMIT AGAIN
IF WE WERE TO TRY TO DO DIRECT SUBSTITUTION
WE END UP WITH THE 0 IN THE DENOMINATOR,
SO THAT'S NOT GOING TO WORK.
SO WHAT WE'RE GONNA DO NOW IS RATIONALIZE THE NUMERATOR
BY MULTIPLYING THE CONJUGATE OF THE NUMERATOR.
SO WE MULTIPLY BOTH THE NUMERATOR AND THE DENOMINATOR
BY THE SQUARE ROOT OF X + DELTA X + THE SQUARE ROOT OF X.
NOW WE NEED TO MULTIPLY THIS NUMERATOR TOGETHER AND SIMPLIFY.
IN THE NUMERATOR THE SQUARE ROOT OF X + DELTA X
x THE SQUARE ROOT OF X + DELTA X WOULD JUST BE X + DELTA X.
SO WHAT WE'RE GONNA DO HERE
IS WE'RE GONNA MULTIPLY THIS x THIS AND THEN THIS x THIS.
BUT NOW THESE ARE CONJUGATES,
SO THE NEXT TWO PRODUCTS WILL BE OPPOSITES.
SO THE SQUARE ROOT OF X + DELTA X x +SQUARE ROOT OF X
AND THEN WE HAVE
THE -SQUARE ROOT OF X x THE SQUARE ROOT OF X + DELTA X,
THOSE ARE OUR OPPOSITES.
SO THOSE WOULD COMBINE TO GIVE US 0
AND THEN WE HAVE -SQUARE ROOT OF X x +SQUARE ROOT OF X
WHICH WOULD JUST GIVE US -X AND THEN IN OUR DENOMINATOR
REMEMBER WE'RE NOT GOING TO MULTIPLY THIS TOGETHER.
WE'RE GONNA LEAVE IT IN FACTORED FORM.
SO IN THE DENOMINATOR WE WOULD HAVE DELTA X
x THE SQUARE ROOT OF X + DELTA X + THE SQUARE ROOT OF X.
LET'S SIMPLIFY.
NOTICE WE HAVE AN X AND THEN A -X WHICH WOULD LEAVE US
WITH DELTA X OVER THE SAME DENOMINATOR
AND NOTICE AGAIN WE HAVE A COMMON FACTOR OF DELTA X.
SO SIMPLIFYING AGAIN WOULD LEAVE US WITH THE LIMIT OF 1
OVER THE SQUARE ROOT OF X + DELTA X + THE SQUARE ROOT OF X.
DELTA X IS APPROACHING 0,
SO WHAT WE'D END UP WITH IS 1 OVER THE SQUARE ROOT OF X
+ THE SQUARE ROOT OF X WHICH IS = 1/2 SQUARE ROOT X.
THEREFORE OUR DERIVATIVE IS = 1/2 x THE SQUARE ROOT OF X.
SO IF THEY WANNA KNOW THE VALUE OF THE DERIVATIVE
AT THE POINT (1,1)
WE NEED TO VALUE WITH A DERIVATIVE AT X = 1
WHICH OF COURSE IS = TO 1/2.
THIS SHOULD BE THE SLOPE OF OUR TANGENT LINE AT THE POINT (1,1).
LET'S VERIFY THIS GRAPHICALLY.
SO HERE'S OUR FUNCTION, HERE'S OUR TANGENT LINE.
OUR POINT WAS (1,1)
AND YOU CAN SEE THE SLOPE OF THE RED LINE, UP 1, OVER 2.
WE HAVE THE CORRECT TANGENT LINE.
WE MIGHT AGAIN WANNA TAKE A MOMENT
AND TAKE A LOOK AT THE GRAPH OF THE DERIVATIVE.
SO AGAIN HERE'S OUR TANGENT LINE, HERE'S OUR FUNCTION,
HERE'S OUR DERIVATIVE FUNCTION.
THE Y VALUES OF THE DERIVATIVE FUNCTION SHOULD GIVE US
THE SLOPES OF THE TANGENT LINES AT THE CORRESPONDING X VALUES.
BY LOOKING AT THE GREEN FUNCTION,
OUR DERIVATIVE FUNCTION,
WE CAN SEE THAT WHEN X = 1 THE Y VALUE LOOKS LIKE IT'S 1/2
WHICH AGAIN IS THE SAME AS THE SLOPE OF THE TANGENT LINE
AT THAT CORRESPONDING X VALUE.
OKAY, LET'S TAKE A LOOK AT 1 MORE.
GIVEN THE FUNCTION F OF X = 1/X
FIND THE EQUATION OF THE TANGENT LINE AT THE POINT (0.5,2)
USING THE LIMIT DEFINITION. LET'S SETUP OUR DERIVATIVE.
SO F OF X + DELTA X WILL BE
1/X + DELTA X - 1/X ALL DIVIDED BY DELTA X.
AGAIN WE'RE NOT ALLOWED TO DIRECT SUBSTITUTION.
WE'D HAVE A 0 IN THE DENOMINATOR.
SO IN THIS CASE WHAT WE'RE GONNA DO
IS MULTIPLY THE NUMERATOR AND DENOMINATOR
BY THE LCD OF THE NUMERATOR.
SO WE'RE GONNA MULTIPLY BOTH THE TOP AND THE BOTTOM
BY X x X + DELTA X.
YOU MIGHT WANNA THINK OF THIS NUMERATOR AS BEING OVER 1
AND THAT MIGHT HELP US MULTIPLE THESE 2 NUMERATORS TOGETHER.
LET'S GO AHEAD AND TRY IT.
WE MULTIPLY THESE FIRST TWO TOGETHER,
NOTICE HOW WE HAVE X + DELTA X ON TOP,
X + DELTA X ON THE BOTTOM.
THOSE ARE GONNA SIMPLIFY OUT
SO THE RESULT WOULD JUST BE X MINUS--ON THE SECOND PRODUCT,
NOTICE THERE'S X ON THE TOP, X ON THE BOTTOM.
THOSE WOULD SIMPLIFY,
SO YOU WOULD HAVE -X + DELTA X.
AGAIN, WE'RE GONNA LEAVE THE DENOMINATOR IN FACTORED FORM
SO WE'RE GONNA HAVE A DELTA X x X x X + DELTA X.
SIMPLIFYING, HERE WE HAVE AN X - AN X.
WE'D HAVE A -DELTA X ALL OVER THE SAME DENOMINATOR.
THE PROBLEM DELTA X SIMPLIFIES OUT AS BEFORE.
THIS WOULD LEAVE A -1/X x (X + DELTA X).
HOWEVER, DELTA X IS APPROACHING 0,
SO YOU CAN SEE WE'D HAVE A -1 IN THE NUMERATOR
AND THEN WE'D HAVE X SQUARED IN THE DENOMINATOR.
THIS IS OUR DERIVATIVE FUNCTION.
SO IF WE WANNA KNOW THE EQUATION OF THE TANGENT LINE
FIRST WE NEED TO FIND THE SLOPE OF THE TANGENT LINE.
SO WE NEED TO EVALUATE THE DERIVATIVE AT 0.5.
I'M GONNA WRITE THAT AS 1/2.
SO WE NEED TO FIND 1--I'M SORRY THERE'S A NEGATIVE HERE,
-1 DIVIDED BY 1/2 SQUARED.
WELL, OF COURSE, 1/2 SQUARED IS 1/4.
SO -1 DIVIDED BY -1/4 WOULD BE TIMES 4.
THAT WOULD GIVE US A SLOPE OF -4.
OKAY. WE HAVE THE POINT OF 0.52 AND WE HAVE A SLOPE OF -4.
LET'S GO AHEAD AND FIND
THE EQUATION OF THAT TANGENT LINE NOW.
USING POINT SLOPE FORM YOU WOULD HAVE
Y - 2 = -4 x (X - 0.5).
REMEMBER THIS IS COMING FROM POINT SLOPE FORM OF THE LINE.
SIMPLIFY THIS AND WE SHOULD HAVE OUR EQUATION OF OUR LINE.
ADDING 2 TO BOTH SIDES WE HAVE Y = -4X + 4.
OKAY, LET'S GO AHEAD AND VERIFY THIS AS WELL.
SO IN BLUE WE HAVE OUR FUNCTION
AND IN RED WE HAVE OUR TANGENT LINE AT THE POINT 0.52.
HERE'S OUR POINT OF TANGENCY
AND WE SAID Y INTERCEPT WAS +4.
OUR SLOPE WAS -4, SO IF I GO DOWN 4, RIGHT 1,
YOU CAN SEE WE'RE STILL ON THE LINE.
THAT VERIFIES OUR WORK IS CORRECT.
AGAIN, LET'S TAKE A LOOK AT THE GRAPH OF THE DERIVATIVE FUNCTION
THE DERIVATIVE FUNCTION IS GRAPHED IN GREEN.
WE CAN SEE THAT WHEN X = 1/2 THAT THE Y VALUE IS -4
AGAIN VERIFYING THAT WE DO HAVE THE CORRECT SLOPE
AS WELL AS THE CORRECT DERIVATIVE.
I SHOULD MENTION THE FOLLOWING ALTERNATIVE LIMIT FORM
OF THE DERIVATIVE IS USEFUL
IF YOU JUST NEED TO FIND THE VALUE OF A DERIVATIVE
AT A GIVEN VALUE X = C.
HOWEVER IF YOU HAVE F PRIME OF X
YOU CAN FIND THE VALUE OF THE DERIVATIVE
AT ANY DEFINED X VALUE YOU WISH.
IF YOU WANT TO YOU CAN GO BACK AND GIVE THIS FORMULA A TRY.
THANK YOU FOR WATCHING.