- 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.
