- WE WANT TO FIND 
THE DERIVATIVE OF F OF X
= 2X SQUARE ROOT X + 4 DIVIDED 
BY X SQUARED SQUARE ROOT X,
AND THEN FIND F PRIME OF 1.
WELL, BEFORE WE FIND 
OUR DERIVATIVE
WE NEED TO RECOGNIZE 
TWO THINGS.
FIRST WE NEED TO REWRITE 
THE SQUARE ROOTS
USING RATIONAL EXPONENTS.
AND THEN SECOND WE NEED 
TO REWRITE THESE TERMS
SO WE HAVE X RAISED 
TO SOME POWER
SO WE CAN APPLY THE POWER RULE 
OF DIFFERENTIATION.
LET'S FIRST WRITE 
THE SQUARE ROOTS
USING RATIONAL EXPONENTS.
TO REVIEW, IF WE HAVE 
THE NTH ROOT OF A TO THE M
THIS IS EQUAL TO A RAISED TO 
THE POWER OF M DIVIDED BY N.
SO USING RATIONAL EXPONENTS, 
THE DENOMINATOR IS THE INDEX
AND THE NUMERATOR 
IS THE EXPONENT.
BECAUSE THESE ARE SQUARE ROOTS 
THE INDEX IS 2, AND THIS IS X,
MEANING THIS IS X 
TO THE FIRST, AND SO IS THIS.
SO WE CAN WRITE THE SQUARE 
ROOT OF X AS X TO THE 1/2.
SO THE ORIGINAL FUNCTION 
CAN BE WRITTEN AS F OF X =
INSTEAD OF 2X SQUARE ROOT X,
WE CAN WRITE THIS 
AS 2X x X TO THE 1/2.
REMEMBER THIS X HERE WOULD 
ALSO HAVE AN EXPONENT OF 1,
+ 4 DIVIDED BY X SQUARED, 
AGAIN, x X TO THE 1/2.
NOW LET'S GO AHEAD AND FIND 
THESE PRODUCTS INVOLVING X.
WHEN MULTIPLYING 
AND THE BASES ARE THE SAME
WE ADD THE EXPONENTS.
SO WE'D HAVE F OF X = 2X 
TO THE POWER OF 1 + 1/2
IS 1 1/2 OR 3/2 + 4 DIVIDED 
BY X TO THE POWER OF 2 + 1/2
THAT WOULD BE 2 1/2 OR 5/2.
SO WE HAVE X TO THE 5/2 
IN THE DENOMINATOR.
SO THIS TERM LOOKS GOOD,
BUT TO APPLY THE POWER RULE 
OF DIFFERENTIATION
WE NEED TO REWRITE THIS TERM 
HERE USING NEGATIVE EXPONENTS.
TO REVIEW THIS PROPERTY,
IF WE HAVE 1 DIVIDED BY A 
RAISED TO THE POWER OF M,
WE CAN REWRITE IT AS A 
RAISED TO THE POWER OF -M.
SO IF THIS CROSSES 
THE FRACTION BAR
IT'S GOING TO CHANGE THE SIGN 
OF THE EXPONENT.
SO IF WE MOVE THIS UP 
TO THE NUMERATOR
IT'S GOING TO CHANGE THE SIGN 
OF THE EXPONENT TO -5/2.
SO NOW WE HAVE F OF X = 2X 
TO THE 3/2 + 4X TO THE -5/2.
AND NOW FINALLY WE CAN FIND 
THE DERIVATIVE
USING OUR POWER RULE 
OF DIFFERENTIATION.
SO LET'S GO AHEAD AND DO THAT.
F PRIME OF X IS GOING TO BE 
EQUAL TO 2
x THE DERIVATIVE OF X 
TO THE 3/2.
SO WE'LL HAVE 2 
x THE EXPONENT OF 3/2
x X TO THE POWER OF 3/2 - 1 
+ 4 x THE EXPONENT OF -5/2
x X TO THE POWER OF -5/2 - 1.
SO THIS IS OUR DERIVATIVE 
FUNCTION,
BUT NOW WE NEED 
TO SIMPLIFY THIS.
THIS IS 2/1 
SO THESE TWO SIMPLIFY NICELY.
SO OUR FIRST TERM WOULD BE 3X 
TO THE POWER OF 3/2 - 1
OR 3/2 - 2/2 THAT'S 1/2.
HERE THE 4 AND THE 2 SIMPLIFY, 
THIS SIMPLIFIES TO 1,
THIS SIMPLIFIES TO 2.
SO WE HAVE + -10 OR - 10, 
X TO THE POWER OF -5/2 - 1
OR -5/2 - 2/2 WOULD BE -7/2.
LET'S GO AHEAD 
AND REWRITE THE SECOND TERM
USING A POSITIVE EXPONENT.
SO, AGAIN, WE CAN THINK 
OF THIS AS BEING OVER ONE.
IF WE MOVE THIS 
DOWN TO THE DENOMINATOR
IT'S GOING TO CHANGE THE SIGN 
OF THE EXPONENT.
SO WE'D HAVE F PRIME OF X 
= 3X TO THE 1/2 - 10
DIVIDED BY X 
TO THE POWER OF 7/2.
NOW, THIS WOULD BE 
OUR DERIVATIVE
USING RATIONAL EXPONENTS,
BUT I ALSO WANT TO SHOW 
HOW TO REWRITE THIS
USING SQUARE ROOTS
SINCE THE ORIGINAL FUNCTION 
WAS GIVEN USING SQUARE ROOTS.
WE WOULD HAVE F PRIME OF X = 
WELL, X TO THE 1/2 IS THE SAME
AS THE SQUARE ROOT OF X.
THIS WOULD BE 3 SQUARE ROOT 
OF X.
NOTICE HOW THE INDEX IS 2
BECAUSE WE HAVE A SQUARE ROOT 
THAT'S THE DENOMINATOR.
AND THE EXPONENT ON THIS X 
IS 1 WHICH IS THE NUMERATOR.
THEN WE'D HAVE MINUS 10.
AND THEN FOR X TO THE 7/2 
THAT'S A LITTLE TRICKY.
7/2 IS THE SAME AS 3 1/2.
SO WE CAN WRITE THIS 
AS 3 + 1/2,
WHICH MEANS IF WE HAVE X 
TO THE 7/2
WE CAN REWRITE THIS AS X 
TO THE THIRD x X TO THE 1/2.
REMEMBER WHEN MULTIPLYING 
WE ADD THE EXPONENTS,
WHICH MEANS YOU CAN WRITE X 
TO THE 7/2
AS X TO THE THIRD 
x THE SQUARE ROOT OF X.
SO, AGAIN, DEPENDING 
ON THE TEXT OR THE INSTRUCTOR,
YOU MAY BE ABLE TO LEAVE 
YOUR DERIVATIVE FUNCTION
IN THIS FORM HERE.
BUT IF WE NEED THE DERIVATIVE 
USING SQUARE ROOTS
IT WOULD BE IN THIS FORM HERE.
THE SECOND PART 
OF THIS QUESTION
ASKED US TO FIND F PRIME OF 1,
WHICH WILL GIVE US THE SLOPE 
OF THE TANGENT LINE
WHEN X = 1.
LET'S FIND THIS 
ON THE NEXT SLIDE.
SO TO FIND F PRIME OF 1 
WE'LL SUBSTITUTE 1 FOR X
INTO THE DERIVATIVE FUNCTION.
WELL, THE SQUARE ROOT OF 1 
IS EQUAL TO 1,
SO WE JUST HAVE 3 - THIS WOULD 
BE 10 DIVIDED BY 1 OR JUST 10.
SO F PRIME OF 1 
IS EQUAL TO -7,
WHICH, AGAIN, WOULD BE 
THE SLOPE WITH A TANGENT LINE
WHEN X = 1.
LET'S GO AHEAD 
AND VERIFY THIS GRAPHICALLY.
THE RED GRAPH IS THE GRAPH 
OF OUR FUNCTION.
NOTICE WHEN X = 1
WE'D BE AT THIS POINT HERE 
ON THE FUNCTION.
THE BLUE LINE WOULD BE 
THE TANGENT LINE WHEN X = 1.
AND SINE F PRIME OF 1 
IS EQUAL TO -7
THE SLOPE WITH THIS 
BLUE TANGENT LINE IS -7.
I HOPE YOU FOUND THIS HELPFUL.
