- GIVEN THE FUNCTION Y, 
WE WANT TO FIND THE Dy/DX
OR THE DERIVATIVE FUNCTION.
BUT LOOKING AT OUR FUNCTION Y
NOTICE HOW WE DEFINITELY HAVE 
A PRODUCT OF TWO FUNCTIONS HERE
BUT ALSO THE FIRST AND SECOND 
FUNCTIONS
ARE ALSO COMPOSITE FUNCTIONS.
SO WE'LL HAVE TO APPLY 
THE PRODUCT RULE
WITH THE CHAIN RULE.
WHENEVER WE'RE WORKING 
WITH THE CHAIN RULE THOUGH,
WE NORMALLY WANT TO BEGIN BY 
IDENTIFYING THE INNER FUNCTION.
SO LOOKING 
AT THIS FIRST FUNCTION,
NOTICE HOW THE INNER FUNCTION 
WOULD BE 3X TO THE 5th + 1
WHICH WE'LL CALL U.
LOOKING AT THE SECOND FUNCTION,
NOTICE HOW THE INNER FUNCTION 
WOULD BE 2X TO THE 7th - 5.
WE'LL CALL THIS V.
LET'S KEEP TRACK OF ALL THIS 
ON THE SIDE.
WE HAVE U = 3X TO THE FIFTH + 1.
BECAUSE WE'LL BE APPLYING 
THE CHAIN RULE,
WE ALSO WANT TO FIND THE U DX 
OR U PRIME.
U PRIME WOULD BE EQUAL 
TO WHAT 15X TO THE 4th,
THE DERIVATIVE OF 1 WOULD BE 0.
AND THEN V WOULD BE EQUAL TO 2X 
TO THE 7th - 5.
SO THE PRIME OR D V D X WOULD BE 
EQUAL TO WHAT 14 X TO THE 6th.
AND AGAIN THE DERIVATIVE 
OF A CONSTANT IS 0.
NOW WE CAN THINK 
OF DIVISIONAL FUNCTION
AS Y = U TO THE 4th x V 
TO THE 9th.
AND NOW WE'LL APPLY THE PRODUCT 
RULE TO FIND D Y D X.
WE'RE LOOKING AT OUR PRODUCT 
RULE THE FIRST FUNCTION F
WOULD ACTUALLY BE U TO THE 4th
AND THE SECOND FUNCTION G 
WOULD BE V TO THE 9th,
WHICH MEANS D Y D X WOULD BE 
EQUAL TO THE FIRST FUNCTION,
U TO THE 4th,
x THE DERIVATIVE OF V TO THE 9th 
WITH RESPECT TO X.
THAT WOULD BE THE DERIVATIVE 
WITH RESPECT TO X OF V
TO THE 9th
WHICH WILL REQUIRE THE CHAIN 
RULE + THE SECOND FUNCTION,
V TO THE 9th x THE DERIVATIVE 
OF THE FIRST FUNCTION
WITH RESPECT TO X.
THE DERIVATIVE OF U TO THE 4th 
WITH RESPECT TO X
WHICH AGAIN WILL REQUIRE 
THE CHAIN RULE.
NOW LET'S FIND THIS DERIVATIVE 
AND THIS DERIVATIVE.
SO WE HAVE D Y D X = U 
TO THE 4th
x THE DERIVATIVE OF V 
TO THE NINTH WITH RESPECT TO X
WOULD BE 9 V TO THE 8th 
x V PRIME.
REMEMBER WE HAVE TO APPLY 
THE CHAIN RULE HERE
BECAUSE WE'RE DIFFERENTIATING 
WITH RESPECT TO X NOT JUST V.
THEN WE HAVE + V TO THE 9th
x THE DERIVATIVE OF U TO THE 4th 
WITH RESPECT TO X,
THAT WOULD BE 4 U TO THE 3rd 
x U PRIME.
AND NOW WE HAVE 
OUR DERIVATIVE FUNCTION
IT'S JUST NOT IN TERMS OF X.
SO NOW WELL MAKE SUBSTITUTIONS 
FOR U, U PRIME, V AND V PRIME.
SO D, Y, D, X IS GOING TO BE 
EQUAL TO U TO THE 4th,
THAT WOULD BE 3 X TO THE FIFTH + 
1 TO THE 4th x 9 x V TO THE 8th
WHICH WOULD BE 2X TO THE 7th - 5 
TO THE 8th x V PRIME
WHICH IS 14 EXIT THE 6,
THEN WE HAVE + TO THE 9th
WHICH WOULD BE 2X TO THE 7th - 5 
TO THE 9th x 4 x U TO THE 3rd
THAT'S 3X TO THE 5th 
+ 1 TO THE 3rd x U PRIME
WHICH WOULD BE 15 X TO THE 4th.
LET'S BEGIN TO SIMPLIFY THIS.
9 x 14X TO THE 6th WOULD BE 126X 
TO THE 6th.
THEN WE HAVE 4 FACTORS OF 3X 
TO THE 5th + 1,
AND WE HAVE 8 FACTORS OF 2X 
TO THE 7th - 5 +
LOOKING AT THE SECOND PRODUCT 
HERE,
WE HAVE 4 x 15 X TO THE 4th 
THAT WOULD BE 60X TO THE 4th.
THEN WE HAVE 9 FACTORS OF 2X 
TO THE 7th - 5
AND 3 FACTORS OF 3X 
TO THE 5th + 1.
SO IN MANY CASES 
YOU CAN PROBABLY LEAVE
YOUR DERIVATIVE FUNCTION 
IN THIS FORM HERE.
BUT -- DERIVATIVES SHOW HOW 
THEY CAN FACTOR THIS FURTHER
AND SIMPLIFY FURTHER.
SO LET'S TAKE THIS 
ON THE NEXT SLIDE AND CONTINUE.
LOOKING AT THIS PRODUCT 
AND THIS PRODUCT,
LET'S BEGIN BY LOOKING AT 126X 
TO THE 6th AND 60X TO THE 4th.
THE GREATEST COMMON FACTOR 
OF THESE TWO TERMS
WE BE 6X TO THE 4th.
SO FOR THE NEXT STEP WE'RE GOING 
TO FACTOR OUT 6X TO THE 4th
BUT ALSO IF WE LOOK AT THE 
FACTORS OF 3X TO THE 5th + 1,
WE HAVE 4 OF THEM HERE 
AND 3 OF THEM THERE.
LET'S GO AHEAD AND FACTOR OUT 
3 OF THEM.
IF WE TAKE A LOOK AT THE FACTORS 
OF 2X TO THE 7th - 5,
WE HAVE 8 OF THEM HERE 
AND 9 OF THEM HERE.
SO WE'LL GO AHEAD 
AND FACTOR OUT 8 OF THEM.
SO LET'S SEE WHAT'S LEFT.
IF WE FACTOR OUT 6X TO THE 45th 
FROM 126X TO THE 6th,
THAT LEAVES US WITH 21X 
TO THE 2nd.
WE'RE FACTORING OUT 3 FACTORS 
OF 3X TO THE 5th + 1
BUT WE HAD 4 OF THEM THAT 
LEAVES US WITH 1 FACTOR OF 3X
TO THE 5th + 1.
AND WE'RE FACTORING OUT 
ALL OF THE FACTORS OF 2X
TO THE 7th - 5.
AND NOW FOR THE 2nd PRODUCT,
WE FACTOR OUT 6X TO THE 4th 
THAT WOULD LEAVE US WITH 10.
WE'RE FACTORING OUT 8 FACTORS 
OF 2X TO THE 7th - 5.
WE HAD 9 OF THEM THOUGH, 
SO THERE'S 1 FACTOR LEFT,
AND NOTICE HOW WE FACTOR OUT ALL 
3 FACTORS OF 3X TO THE 5th + 1.
SO NOW WE'LL SIMPLIFY INSIDE 
THESE BRACKETS BY DISTRIBUTING
AND COMBINING LIKE TERMS.
SO WE'D HAVE 63X TO THE 7th 
+ 21X SQUARED
AND THEN + 20X TO THE 7th - 50
AND FOR THE LAST STEP 
WE DO HAVE LIKE TERMS HERE.
THERE ARE 2 X TO THE 7th TERMS.
SO IF YOU ARE ASKED TO FACTOR 
THE DERIVATIVE FUNCTION
COMPLETELY,
IT WOULD LOOK LIKE THIS.
WE'D HAVE 6X TO THE 4th 
x 3X TO THE FIFTH
+ 1 TO THE 3rd x 2X TO THE 7th 
- TO THE 8th
AND THEN WE'D HAVE 83 X 
TO THE 7th + 21X SQUARED - 50.
PERSONALLY I DON'T THINK 
ALL THIS ALGEBRA IS NECESSARY
BUT JUST IN CASE 
YOU ARE REQUIRED TO SHOW IT,
HERE'S THE FACTORED FORM 
OF A DERIVATIVE FUNCTION.
BUT IF WE'RE NOT REQUIRED 
TO SIMPLIFY IT AS MUCH AS THIS,
WE COULD JUST LEAVE IT 
IN THIS FORM HERE.
I HOPE YOU FOUND THIS HELPFUL.
