- TO DETERMINE THE DERIVATIVE 
OF THE GIVEN FUNCTION,
WE NEED TO RECOGNIZE TWO THINGS.
FIRST WE HAVE A PRODUCT 
OF TWO FUNCTIONS
SO WE'LL HAVE TO APPLY 
THE PRODUCT RULE.
SO WE'LL LET THE FIRST FUNCTION 
EQUAL 6X SQUARED,
WE'LL CALL IT F.
AND LET THE SECOND FUNCTION BE
THE SQUARE ROOT OF 2X SQUARED 
- 1 WHICH WE'LL LET EQUAL G.
THE NEXT THING 
WE NEED TO RECOGNIZE IS THAT
G IS A COMPOSITE FUNCTION
SO WHEN DIFFERENTIATING G WE'LL 
HAVE TO APPLY THE CHAIN RULE,
BUT WE ALSO DON'T WANT TO LEAVE 
THIS IN MEDICAL FORM.
WE WANT TO RE-WRITE THIS 
USING A RATIONAL EXPONENT.
SO THIS IS THE SAME AS 
6X SQUARED
x THE QUANTITY 2X SQUARED - 1 
TO THE 1/2 POWER.
REMEMBER THIS QUANTITY HERE
WE CAN THINK OF AS BEING RAISED 
TO THE POWER OF 1
AND THE INDEX IS 2,
GIVING US THE RATIONAL EXPONENT 
OF 1/2.
NOW THE NEXT STEP WE'RE GOING TO 
WRITE OUT THE PRODUCT RULE
BUT WE'RE NOT ACTUALLY GOING TO 
DETERMINE G PRIME AND F PRIME.
WE'LL DO THAT 
IN THE FOLLOWING STEP.
SO APPLYING THE PRODUCT RULE 
F PRIME OF X
IS GOING TO BE EQUAL TO THE 
FIRST FUNCTION, 6X SQUARED
x THE DERIVATIVE 
OF THE SECOND FUNCTION.
SO WE'LL HAVE DERIVATIVE 
OF THE QUANTITY
2X SQUARED - 1 
TO THE POWER OF 1/2
+ THE SECOND FUNCTION
x THE DERIVATIVE OF THE FIRST 
FUNCTION, WHICH IS 6X SQUARED.
AND NOW LET'S FOCUS 
ON DETERMINING
THESE TWO DERIVATIVES.
THIS FUNCTION HERE 
IS A COMPOSITE FUNCTION
SO WE HAVE TO APPLY 
THE CHAIN RULE.
SO WE WANT TO LET THE INNER 
FUNCTION EQUAL THE U.
SO IF WE LET 2X SQUARED - 1 = U,
THEN THIS ENTIRE FUNCTION 
WOULD BE U TO THE 1/2.
SO TO DIFFERENTIATE U TO THE 1/2 
WITH RESPECTS TO X,
WE'LL FIND THE DERIVATIVE 
WITH RESPECTS TO U
AND THEN MULTIPLY IT BY U PRIME
WHICH IS GIVEN HERE 
BY THE EXTENDED POWER RULE
WHICH INCLUDES THE CHAIN RULE.
SO F PRIME OF X EQUALS 
6X SQUARED
x THE DERIVATIVE OF U TO THE 1/2  
WOULD BE 1/2,
U TO THE 1/2 - 1 WHICH IS -1/2.
AND AGAIN U IS 2X SQUARED - 1 
x U PRIME
AND SINCE U IS 2X SQUARED - 1, 
U PRIME WOULD BE 4X
+ 2X SQUARED - 1 TO THE 1/2 
x THE DERIVATIVE OF 6X SQUARED,
WHICH WOULD BE 12X.
NOW THAT WE HAVE FOUND 
OUR DERIVATIVES
THE REST OF THIS IS GOING TO BE 
ALGEBRA TO CLEAN THIS UP.
SO FOR THIS FIRST PRODUCT 
WE WOULD HAVE 6X SQUARED x 1/2
THAT'S 3X SQUARED,
x 4X THAT WOULD BE 12X CUBED.
THAT WOULD BE IN THE NUMERATOR
AND THEN FOR THIS
NEGATIVE EXPONENT,
WE'LL MOVE IT DOWN 
TO THE DENOMINATOR
TO MAKE THIS AN EXPONENT 
OF +1/2.
SO AGAIN OUR NUMERATOR 
IS 12X CUBED.
THE DENOMINATOR WOULD BE THE 
QUANTITY 2X SQUARED - 1
TO THE POWER OF 1/2,
PLUS THE SECOND PRODUCT 
WHICH WOULD JUST BE 12X
x THE QUANTITY 2X SQUARED - 1 
TO THE 1/2 POWER.
SO AGAIN THIS IS OUR DERIVATIVE 
FUNCTION
BUT WHAT HAPPENS 
IN MOST TEXTBOOKS
IS THEY DETERMINE THIS SUM.
SO TO DETERMINE THIS SUM,
WE'D HAVE TO THINK OF THIS 
AS THE DENOMINATOR OF 1,
THEN WE'D HAVE TO FIND 
A COMMON DENOMINATOR
TO ADD THESE 2 FRACTIONS.
SO I'LL GO AHEAD AND SHOW THAT 
ON THE NEXT SLIDE.
SO AGAIN THIS WOULD BE 
THE COMMON DENOMINATOR
SO WE'RE GOING TO MULTIPLY 
THIS FRACTION
BY 2X SQUARED - 1 TO THE 1/2/2X 
SQUARED - 1 TO THE 1/2.
SO LET'S SEE WHAT THIS GIVES US.
WE NOW HAVE A COMMON DENOMINATOR 
SO HERE WE HAVE 12X CUBED.
IN THIS FRACTION WE MULTIPLY 
THE QUANTITY
2X SQUARED - 1 TO THE 1/2 
x 2X SQUARED - 1 TO THE 1/2.
NOTICE HOW THE BASES 
ARE THE SAME,
SO WE ADD THE EXPONENTS
WHICH JUST GIVES US A FACTOR 
OF 2X SQUARED - 1.
SO WE END UP WITH + 12X x 2X 
SQUARED - 1 TO THE FIRST POWER
AND OF COURSE WE CAN LEAVE OFF 
THE EXPONENT.
BUT NOW WHAT WE CAN DO 
IS DISTRIBUTE THE 12X.
WHEN WE DISTRIBUTE THIS 12X 
WE'RE GOING TO HAVE 24X CUBED
WHICH WE CAN ADD TO 12X CUBED
SO THAT WILL BE 36X CUBED.
AND THEN - 12X.
AND LET'S GO AHEAD 
AND WRITE THIS RATIONAL EXPONENT
AS A RADICAL
SO WE'D HAVE THE SQUARE ROOT 
OF 2X SQUARED - 1.
THEN THE VERY LAST STEP 
IS TO FACTOR THE NUMERATOR.
THE GREATEST COMMON FACTOR OF 
36X CUBED AND 12X WOULD BE 12X.
SO THE FINAL FORM OF OUR 
NUMERATOR IF WE FACTOR OUT 12X
WOULD BE 12X x 3X SQUARED - 1
ALL OVER THE SQUARE ROOT 
OF 2X SQUARED - 1.
SO A LOT OF THESE TYPES 
OF PROBLEMS END UP BEING ALGEBRA
MORE THAN CALCULUS.
THIS WOULD BE THE ANSWER 
THAT YOU WOULD PROBABLY FIND
IN MOST TEXTBOOKS.
SO I HOPE YOU FOUND THIS EXAMPLE 
HELPFUL
AND IF NOTHING ELSE 
A GOOD REVIEW OF ALGEBRA.
