- NOW WE'LL LOOK AT AN EXAMPLE OF FINDING THE EQUATION
OF A TANGENT LINE USING THE PRODUCT RULE.
SO GIVEN OUR FUNCTION HERE,
WE WANT TO FIND THE EQUATION OF THE TANGENT LINE AT X = -2.
SO BEFORE WE FIND THE DERIVATIVE
IN ORDER TO DETERMINE THE SLOPE OF THE TANGENT LINE AT X = -2,
WE NEED TO RECOGNIZE THAT
IN ORDER TO FIND THE EQUATION OF A LINE,
WE ACTUALLY HAVE TO FIND THE POINT ON THE FUNCTION.
SO THEY'RE GIVING US THE X COORDINATE
BUT WE ALSO HAVE TO FIND THE Y COORDINATE
IN ORDER TO FIND THE EQUATION ON THE LINE.
SO WE KNOW OUR POINT, LET'S CALL IT POINT P,
HAS AN X COORDINATE OF -2 BUT IN ORDER TO FIND THE Y COORDINATE,
WE NEED TO FIND F OF -2.
SO NOW WE'LL SUBSTITUTE -2 FOR X IN OUR FUNCTION.
WE HAVE -2 AND WE'LL HAVE PLUS
THIS WILL BE 2 DIVIDED BY -2 THAT WILL BE -1 x X SQUARED -2
SQUARED IS 4, 4 - 3.
SO WE HAVE -3 X 1 SO THE Y COORDINATE OF OUR POINT IS -3.
NOW WE CAN GO AHEAD AND FIND THE DERIVATIVE
IN ORDER TO FIND THE SLOPE OF THE TANGENT LINE WHEN X = -2.
NOW WE COULD MULTIPLY THIS OUT,
BUT WE'LL GO AHEAD AND APPLY THE PRODUCT RULE.
SO WE'LL CALL THIS FIRST FUNCTION F
AND THE SECOND FUNCTION G.
SO F PRIME OF X IS EQUAL TO THE FIRST FUNCTION WHICH IS X +,
I'M GOING TO GO AHEAD AND WRITE THIS AS 2 x X TO THE -1,
AND IF WE MOVE THIS UP INTO THE NUMERATOR IT'S GOING
TO CHANGE THE SIGN OF THE EXPONENT.
SO THAT'S 2X TO THE -1 x THE DERIVATIVE
OF THE SECOND FUNCTION WHICH WILL BE THE DERIVATIVE
OF X SQUARED - 3.
WE'LL COME BACK AND FIND THAT,
PLUS THE SECOND FUNCTION x THE DERIVATIVE OF THE FIRST FUNCTION
WHICH AGAIN I'M GOING TO WRITE THIS AS X + 2 x X TO THE -1.
NOW WE'LL FIND THE DERIVATIVE HERE AND HERE,
THEN MULTIPLY AND SIMPLIFY.
SO F PRIME OF X IS EQUAL TO X + 2X TO THE -1 x THE DERIVATIVE
OF X SQUARED - 3 IS JUST 2X.
+ THE QUANTITY X SQUARED -3 x THE DERIVATIVE OF X + 2X
TO THE -1, THAT'S GOING TO BE 1
AND THEN FOR THE DERIVATIVE OF 2 X TO -1
THEN MULTIPLY BY -1 SO IT'S GOING TO BE NEGATIVE
OR - 2 X TO THE, SUBTRACT 1 FROM THE EXPONENT
THAT'S GOING TO BE -2.
NOW LET'S GO AHEAD AND FIND THESE PRODUCTS
AND SIMPLIFY OUR DERIVATIVE FUNCTION.
WE HAVE 2X x X THAT'S 2X SQUARED,
AND HERE WE HAVE 2X x 2X TO THE -1, WELL 2 x 2 IS 4,
X TO THE FIRST x X TO THE -1 WOULD BE X TO THE 0,
WHICH IS EQUAL TO 1.
SO THIS PRODUCT IS JUST 4, SO WE'LL HAVE + 4.
HERE WE'RE GOING TO HAVE 4 PRODUCTS
SO WE'LL HAVE X SQUARED x 1 THAT'S X SQUARED.
X SQUARED x -2X TO -2.
AGAIN WE'RE GOING TO ADD THE EXPONENTS ON X
SO X TO THE 0 IS EQUAL TO 1, SO THIS IS JUST -2 OR - 2.
THEN WE HAVE -3 x 1, THAT'S - 3.
THEN WE HAVE -3 x -2X TO THE -2 THAT'S 6X TO THE -2
OR + 6 DIVIDED BY X SQUARED.
IF WE MOVE X TO THE -2 TO THE DENOMINATOR,
IT'S GOING TO CHANGE US OUT OF THE EXPONENT.
SO OUR DERIVATIVE FUNCTION IS EQUAL TO COMBINING LIKE TERMS
WE HAVE 2X SQUARED TERMS AND 3 CONSTANTS.
SO WE HAVE 2X SQUARED + 1X SQUARED
THAT'S GOING TO BE 3X SQUARED AND THEN 4 - 2 - 3,
THAT'S -1 OR - 1 + 6 DIVIDED BY X TO THE SECOND.
AND NOW IF WE EVALUATE THE DERIVATIVE AT X = -2,
IT WILL TELL US THE SLOPE OF THE TANGENT LINE AT THIS POINT.
SO WE'LL SUBSTITUTE -2 FOR X AND THEN EVALUATE THIS.
SO 4 X 3, THAT'S 12, - 1,
THAT'S 11 + THIS IS GOING TO BE 6/4 OR 3 HALVES.
WELL 3 HALVES IS THE SAME AS 1.5 SO 11 + 1.5 IS 12.5
SO THE SLOPE OF OUR LINE IS EQUAL TO 12.5
OR IF WE WANT TO PUT THIS IN FRACTION FORM,
12.5 IS EQUAL TO 12 AND 1/2 WHICH AS A IMPROPER FRACTION
WOULD BE 25 HALVES.
SO NOW THAT WE HAVE THE SLOPE OF OUR TANGENT LINE,
AND A POINT ON THE LINE,
WE CAN NOW FIND THE EQUATION OF OUR TANGENT LINE.
SO TO FIND OUR EQUATION
LET'S GO AHEAD AND USE POINT SLOPE FORM GIVEN HERE
AND TO DO THIS AGAIN WE'RE GOING TO USE THE POINT, -2, -3
AND THE SLOPE OF 25 HALVES.
SO WE'LL HAVE Y - -3 = 25 HALVES x THE QUANTITY X - -2.
GO AHEAD AND SIMPLIFY THIS, WE HAVE Y + 3 = 25 HALVES x X + 2.
LET'S GO AHEAD AND PUT THIS IN SLOPED INTERCEPT FORM
SO WE'LL SOLVE THIS FOR Y AND DISTRIBUTE HERE.
SO WE HAVE Y + 3 = DISTRIBUTING HERE WE'LL HAVE 25 HALVES X,
DISTRIBUTING HERE THAT'S THE SAME AS 2/1.
THE 2s SIMPLIFY OUT SO WE HAVE + 25
AND THEN WE'LL SUBTRACT 3 ON BOTH SIDES.
SO WE HAVE Y = 25 HALVES X + 22.
THIS WOULD BE THE EQUATION OF OUR TANGENT LINE
AT THE POINT -2, -3.
LET'S GO AHEAD AND VERIFY THIS GRAPHICALLY.
THE BLUE GRAPH IS OUR ORIGINAL FUNCTION
THE POINT -2, -3 IS HERE.
NOTICE HOW THE GRAPH OF OUR LINE IS TANGENT AT THAT POINT.
OKAY, I HOPE YOU FOUND THIS HELPFUL.