- WE'RE GIVEN F OF X
AND ASKED TO FIND F PRIME OF X 
AND F PRIME OF 4,
AND NOTICE THE GIVEN FUNCTION 
F OF X IS A QUOTIENT.
SO TO FIND THE DERIVATIVE 
FUNCTION,
WE WILL HAVE TO APPLY THE 
QUOTIENT RULE STATED HERE BELOW,
WHERE THE DERIVATIVE OF FUNCTION 
F DIVIDED BY FUNCTION G
WITH RESPECT TO X
IS EQUAL TO G x F PRIME - F x G 
PRIME DIVIDED BY G SQUARED.
BEFORE WE APPLY 
THE QUOTIENT RULE, THOUGH,
WE DO WANT TO REWRITE 
THESE SQUARE ROOTS
AS RATIONAL EXPONENTS.
AND SINCE FOR A SQUARE ROOT 
THE INDEX IS TWO,
AND SINCE WE HAVE THE SQUARE 
ROOT OF X,
THIS WOULD BE X TO THE FIRST,
THE SQUARE ROOT OF X IS EQUAL TO 
X TO THE 1/2
USING THIS PROPERTY HERE.
SO THE GIVEN FUNCTION, F OF X 
CAN BE WRITTEN AS
2 x X TO THE 1/2 POWER - 1 
DIVIDED BY 2X TO THE 1/2 + 1.
AND NOW, WHEN APPLYING THE 
QUOTIENT RULE,
NOTICE AT OUR FUNCTION F
THE NUMERATOR WOULD BE 
2X TO THE 1/2 -1,
AND THE DENOMINATOR OR G 
WOULD BE 2X TO THE 1/2 + 1.
SO F PRIME OF X WOULD BE EQUAL 
TO THE DENOMINATOR,
WHICH IS 2X TO THE 1/2 + 1 x THE 
DERIVATIVE OF THE NUMERATOR,
THE DERIVATIVE OF 2X TO THE 1/2 
- 1,
AND THEN MINUS THE NUMERATOR,
2X TO THE 1/2 - 1 x THE 
DERIVATIVE OF THE DENOMINATOR,
THE DERIVATIVE OF 2X TO THE 1/2 
+ 1
ALL OVER THE DENOMINATOR 
SQUARED,
WHICH WOULD BE 2X TO THE 1/2 + 1 
SQUARED.
SO NOW, WE'LL FIND 
THE DERIVATIVE HERE,
THE DERIVATIVE HERE 
AND THEN SIMPLIFY.
SO F PRIME OF X IS EQUAL TO--
WELL, THE DENOMINATOR IS GOING 
TO STAY THE SAME FOR RIGHT NOW.
SO WE HAVE THE QUANTITY 
2X TO THE 1/2 + 1
x THE DERIVATIVE OF 2X 
TO THE 1/2 - 1.
THE DERIVATIVE OF 2 x X TO THE 
1/2 WE'D MULTIPLY BY 1/2.
2 x 1/2 = 1 AND THEN SUBTRACT 1 
FROM THE EXPONENT.
1/2 - 1 = -1/2.
SO THIS WOULD BE X TO THE -1/2,
AND THE DERIVATIVE 
OF A CONSTANT WOULD BE ZERO.
SO THEN WE JUST HAVE MINUS 
THE QUANTITY 2X TO THE 1/2 -1
x THE DERIVATIVE 
OF 2X TO THE 1/2 + 1,
AND IF WE MULTIPLY BY 1/2, 
THAT WOULD GIVE US 1.
SUBTRACT ONE FROM THE EXPONENT.
ONCE AGAIN, WE HAVE X TO THE 
POWER OF -1/2,
AND AGAIN, THE DERIVATIVE OF A 
CONSTANT IS ZERO.
SO NOW, WE'LL DISTRIBUTE 
THE X TO THE -1/2 HERE
AS WELL AS HERE.
WHEN WE MULTIPLY 2X TO THE 1/2 x 
X TO THE -1/2,
BECAUSE THE BASES ARE SAME, 
WE ADD THE EXPONENTS.
THAT WOULD BE X TO THE ZERO, 
WHICH IS EQUAL TO ONE.
SO THAT JUST GIVES US TWO, 
AND THEN WE HAVE PLUS--
THIS WOULD BE X TO THE POWER 
OF -1/2,
AND THEN WE HAVE MINUS THE SAME 
THING HERE.
WE MULTIPLY 2X TO THE 1/2 x X TO 
THE -1/2.
THAT'D BE 2X TO THE ZERO 
OR JUST TWO,
AND THEN WE HAVE MINUS X 
TO THE -1/2.
NOW, LET'S GO AHEAD AND SIMPLIFY 
THIS,
AND NOTICE HOW THESE TWOS WOULD 
SIMPLIFY OUT.
2 - 2 = 0,
AND THEN WE HAVE MINUS A 
NEGATIVE, WHICH BECOMES PLUS.
SO THE NUMERATOR BECOMES JUST 2X 
TO THE -1/2,
AND THEN WE HAVE DIVIDED 
BY 2X TO THE 1/2 + 1 SQUARED.
LET'S GO AHEAD AND WRITE THIS 
USING RADICALS, THOUGH.
X TO THE -1/2 IS THE SAME AS 1/X 
TO THE 1/2,
WHICH WOULD BE THE SAME AS ONE 
OVER THE SQUARE ROOT OF X.
SO THIS WOULD BE 2 DIVIDED 
BY THE SQUARE ROOT OF X
DIVIDED BY--AGAIN, X TO THE 1/2,
WHICH WOULD JUST BE THE SQUARE 
ROOT OF X.
SO WE HAVE 2 x THE SQUARE ROOT 
OF X + 1 SQUARED.
SO THIS WOULD BE OUR DERIVATIVE 
FUNCTION,
AND NOW THAT WE HAVE 
THE DERIVATIVE FUNCTION,
WE CAN DETERMINE F PRIME OF 4.
BUT BEFORE WE DO THAT,
LET'S JUST MAKE SURE WE DIDN'T 
LOSE ANYBODY
HERE IN THE NUMERATOR.
IF WE HAVE 2X TO THE POWER 
OF - 1/2,
WE CAN WRITE THAT AS 2 DIVIDED 
BY X TO THE 1/2.
IF WE MOVE THIS 
TO THE DENOMINATOR,
IT CHANGES THE SIGN 
OF THE EXPONENT.
X TO THE 1/2 IS THE SAME 
AS THE SQUARE ROOT OF X.
NOW, LET'S GO AHEAD 
AND EVALUATE THIS AT X = 4.
SO IF WE SUBSTITUTE 4 FOR X,
WE WOULD HAVE 2 DIVIDED 
BY THE SQUARE ROOT OF 4
DIVIDED THE QUANTITY 2 x THE 
SQUARE ROOT OF 4 + 1 SQUARED.
WELL, WE KNOW THE SQUARE ROOT 
OF FOUR IS EQUAL TO TWO,
SO THE NUMERATOR SIMPLIFIES TO 
2/2 OR JUST ONE.
OUR DENOMINATOR WOULD BE 2 x 2. 
THAT'S 4 + 1.
SO WE'D HAVE FIVE SQUARED.
SO IT'S EQUAL TO 1/25.
SO F PRIME OF 4 = 1/25,
WHICH MEANS IF WE GRAPH THE 
ORIGINAL FUNCTION, AT X = 4
THE SLOPE OF THE TANGENT LINE 
WOULD BE 1/25.
LET'S TAKE A LOOK AT THIS.
HERE'S THE GRAPH OF OUR 
FUNCTION.
HERE'S THE POINT OF THE FUNCTION 
WHEN X = 4,
AND SINCE F PRIME OF 4 = 1/25,
THE SLOPE OF THIS TANGENT LINE 
IS 1/25.
I HOPE YOU FOUND THIS HELPFUL.
