- 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.