- WE WANT TO DETERMINE THE DERIVATIVE OF THE GIVEN FUNCTION
USING THE LIMIT DEFINITION OF THE DERIVATIVE
AND THEN DETERMINE F PRIME OF 3
WHICH WILL TELL US THE SLOPE OF THE TANGENT LINE
AND X = 3 ON THE FUNCTION.
SO HERE IS THE LIMIT DEFINITION OF THE DERIVATIVE,
SO F PRIME OF X OR THE DERIVATIVE OF F
IS GOING TO BE EQUAL TO THE LIMIT AS H APPROACHES 0
OF THIS DIFFERENCE QUOTIENT.
NOTICE OUR NUMERATOR IS F OF X + H - F OF X,
SO FOR F OF X + H WE'RE GOING TO REPLACE X WITH X + H.
SO WE'LL HAVE THE SQUARE ROOT OF X + H + 1 - F OF X
WHICH IS JUST THE SQUARE ROOT OF X + 1, ALL OVER H.
SO F OF X + H = THE SQUARE ROOT OF X + H + 1
AND F OF X = THE SQUARE ROOT OF X + 1.
NOW LOOKING AT THIS DIFFERENCE QUOTIENT,
WE WANT TO MANIPULATE THIS
SO THAT WE CAN DETERMINE THIS LIMIT BY DIRECT SUBSTITUTION.
NOTICE IF WE ATTEMPT DIRECT SUBSTITUTION NOW,
WE HAVE DIVISION BY 0 WHICH IS UNDEFINED.
SO WHAT WE'LL DO HERE
IS MULTIPLY THE NUMERATOR AND DENOMINATOR
BY THE CONJUGATE OF THE NUMERATOR,
SO WE'RE GOING TO MULTIPLY BY THE SQUARE ROOT OF X + H + 1
+ THE SQUARE ROOT OF X + 1.
OUR DENOMINATOR IS GOING TO BE THE SAME.
NOW LET'S SEE WHAT THIS GIVES US.
WE'RE GOING TO LEAVE THE DENOMINATOR IN FACTORED FORM,
SO WE'LL HAVE H x SQUARE ROOT OF X + H + 1
+ SQUARE ROOT OF X + 1.
NOW WE MULTIPLY THE NUMERATOR.
BECAUSE THEY'RE CONJUGATES THIS IS GOING TO SIMPLIFY NICELY.
NOTICE FOR THIS PRODUCT WE'RE MULTIPLYING TWO SQUARE ROOTS
WITH THE SAME RADICAND,
SO THE RESULTS WILL JUST BE THE RADICAND OF X + H + 1.
NOW THE NEXT TWO PRODUCTS WILL BE OPPOSITES.
HERE WE HAVE A POSITIVE PRODUCT OF THE SAME TWO SQUARE ROOTS,
SO THAT WILL BE 0.
AND THEN WE'RE GOING TO HAVE MINUS THE PRODUCT
OF THE SQUARE ROOT OF X + 1 x THE SQUARE ROOT OF X + 1
WHICH IS JUST X + 1,
BUT WE ARE SUBTRACTING (X + 1).
NOW, LET'S GO AHEAD AND CLEAR THESE PARENTHESES
AND SEE WHAT SIMPLIFIES.
THERE SHOULD BE A CLOSED PARENTHESIS HERE AS WELL.
SO WE HAVE X + H + 1,
AND THEN -X - 1 ALL OVER THE SAME DENOMINATOR.
AND NOW WE ALMOST HAVE IT HERE.
NOTICE THAT WE HAVE X - X. THAT'S 0.
AND WE HAVE 1 - 1, THAT'S 0,
SO LET'S GO AHEAD AND WRITE THIS ONE MORE TIME.
OUR NUMERATOR IS JUST H,
AND WE HAVE A FACTOR OF H IN THE DENOMINATOR,
SO THAT'S GOING TO SIMPLIFY OUT.
IF AGAIN, WE HAVE A COMMON FACTOR OF H HERE THAT'S = TO 1,
AND NOW AS H APPROACHES 0 OUR DENOMINATOR
IS GOING TO APPROACH THE SQUARE ROOT OF X + 1
+ THE SQUARE ROOT OF X + 1 WHICH = 2 SQUARE ROOT OF X + 1.
SO OUR NUMERATOR IS 1,
AND OUR DENOMINATOR IS 2 SQUARE ROOT OF X + 1.
AND THIS IS THE DERIVATIVE OF OUR GIVEN FUNCTION.
SO NOW THAT WE HAVE THE DERIVATIVE
WE CAN DETERMINE THE VALUE OF THE DERIVATIVE AT X = 3.
LET'S GO AHEAD AND DO THAT ON THE NEXT PAGE.
SO IF WE REPLACE X WITH 3 IN THE DERIVATIVE,
WE'LL HAVE 1/2 x THE SQUARE ROOT OF 3 + 1,
THAT'S GOING TO BE 4.
SO I HAVE 1/2 x 2 WHICH = 1/4.
SO THIS TELLS US THE SLOPE OF THE TANGENT LINE
TO OUR FUNCTION IS 1/4 AT X = 3.
REMEMBER, OUR FUNCTION WAS F OF X = THE SQUARE ROOT OF X + 1.
SO IF WE WANTED TO KNOW THE POINT OF TANGENCY
OR THE POINT WHERE THE SLOPE OF THE TANGENT LINE = 3,
WE'D HAVE TO EVALUATE THE FUNCTION AT X = 3.
SO IT DOESN'T ASK US IN THE QUESTION,
BUT THE POINT OF TANGENCY WOULD BE (3,2),
AND LET'S GO AHEAD AND VERIFY THIS GRAPHICALLY.
HERE'S OUR FUNCTION F OF X,
SO IN RED WE HAVE THE GRAPH OF OUR FUNCTION
WHICH IS F OF X = THE SQUARE ROOT OF X + 1.
NOTICE HOW OUR POINT OF TANGENCY IS THE POINT (3,2)
AND FROM THE DERIVATIVE WE DETERMINE
THAT THE SLOPE OF THIS TANGENT LINE = 1/4.
AND THAT'S GOING TO DO IT FOR THIS EXAMPLE.
I HOPE YOU FOUND THIS HELPFUL.