- IN THIS EXAMPLE WE'RE ASKED TO USE LOGARITHMIC DIFFERENTIATION
IN ORDER TO FIND THE DERIVATIVE OF F OF X = THE QUANTITY 2X - 3
SQUARED x THE QUANTITY 5X
SQUARED + 2 CUBED. LET'S START WITH A QUICK REVIEW ON HOW TO
PERFORM LOG DIFFERENTIATION. STEP ONE; WE'LL TAKE THE
NATURAL LOG OF BOTH SIDES OF THE EQUATION. STEP TWO; WE'LL
APPLY THE PROPERTIES OF LOGARITHMS TO EXPAND THE
LOGS AS MUCH AS POSSIBLE. STEP THREE; WE'LL DIFFERENTIATE BOTH
SIDES OF THE EQUATION. STEP FOUR,
WE'LL SOLVE FOR DY/DX AND THEN STEP FIVE,
WE'LL USE THE ORIGINAL FUNCTION AS SUBSTITUTE FOR X
INTO OUR DERIVATIVE FUNCTION.
BEFORE WE START, WE SHOULD NOTE
THAT WE DON'T HAVE TO USE LOG DIFFERENTIATION
TO FIND THIS DERIVATIVE, WE COULD APPLY THE PRODUCT RULE
COMBINED WITH THE CHAIN RULE IN ORDER
TO FIND OUR DERIVATIVE FUNCTION. BUT FOR THIS EXAMPLE
WE WILL USE LOG DIFFERENTIATION. LET'S FIRST
REPLACE F OF X WITH Y SO WE WOULD HAVE Y = THE QUANTITY 2X
- 3 SQUARED x THE QUANTITY 5X SQUARED + 2 CUBED AND NOW WE'LL
TAKE THE NATURAL LOG OF BOTH SIDES OF THE EQUATION
AND NOW WE'LL START TO EXPAND THE RIGHT SIDE OF
THIS EQUATION USING OUR PROPERTY OF LOGARITHMS GIVEN
HERE BELOW. SO BECAUSE WE HAVE A PRODUCT HERE WE CAN BREAK
THIS UP INTO A SUM OF TWO LOGS. WE WOULD HAVE NATURAL LOG Y =
NATURAL LOG OF THE QUANTITY 2X - 3 SQUARED + NATURAL LOG
OF THE QUANTITY 5X SQUARED + 2 CUBED, AND NOW WE CAN APPLY THE
POWER PROPERTY OF LOGARITHMS WHICH MEANS WE CAN TAKE THIS
EXPONENT HERE AND MOVE IT TO THE FRONT SO IT'S THE
COEFFICIENT AND WE CAN DO THE SAME WITH THIS 3 HERE. SO NOW
WE HAVE NATURAL LOG Y = 2 x NATURAL LOG OF THE QUANTITY 2X
- 3 + 3 x NATURAL LOG OF THE QUANTITY 5X SQUARED + 2.
NOW THAT WE'VE EXPANDED THIS AS MUCH AS POSSIBLE WE'RE NOW
GOING TO DIFFERENTIATE BOTH SIDES
OF THE EQUATION WITH RESPECT TO X.
NOW WE'LL DIFFERENTIATE
BOTH SIDES OF THE EQUATION WITH RESPECT TO X AND TO DO
THIS WE'LL BE USING THE DERIVATIVE FORMULA GIVEN HERE.
THE DERIVATIVE OF NATURAL LOG U WITH RESPECT
TO X IS EQUAL TO 1/U x U PRIME. AGAIN, NOTICE
HOW THIS IS THE CHAIN RULE WHERE THIS FIRST FACTOR IS A
DERIVATIVE OF THE OUTER FUNCTION AND THE SECOND FACTOR,
U PRIME IS A DERIVATIVE OF THE INNER FUNCTION.
BUT ALSO, TAKING A LOOK AT THIS FIRST DERIVATIVE HERE
SINCE OUR FUNCTION IS IN TERMS OF Y THIS IS IMPLICIT
DIFFERENTIATION SO WE'LL HAVE AN EXTRA FACTOR OF DY/DX WHICH
IS ACTUALLY JUST THE CHAIN RULE. SO THE DERIVATIVE
OF NATURAL LOG Y WITH RESPECT TO X WOULD BE 1/Y x DY/DX.
AGAIN, THIS REQUIRES IMPLICIT DIFFERENTIATION BECAUSE OUR
FUNCTION IS IN TERMS OF Y AND WE'RE DIFFERENTIATING
WITH RESPECT TO X, AND NOW ON THE RIGHT SIDE WE'LL HAVE 2
x THE DERIVATIVE OF NATURAL LOG 2X - 3 SO OUR
INNER FUNCTION U IS EQUAL TO 2X - 3. SO AGAIN, THE DERIVATIVE
WOULD BE 1/U x U PRIME OR 1/THE QUANTITY 2X - 3 x THE
DERIVATIVE OF 2X - 3 WITH RESPECT
TO X WHICH WOULD BE 2 + 3 x THE DERIVATIVE OF NATURAL LOG
OF THE QUANTITY 5X SQUARED + 2. AGAIN, WE HAVE TO APPLY THE
CHAIN RULE WHERE THE INNER FUNCTION OR U IS EQUAL
TO 5X SQUARED + 2. SO OUR DERIVATIVE IS EQUAL TO 1/U x U
PRIME OR 1/THE QUANTITY 5X SQUARED + 2 x THE DERIVATIVE
OF 5X + 2 WITH RESPECT TO X WHICH WOULD BE 10X. NOW THAT
WE'VE DIFFERENTIATED WE HAVE TO SIMPLIFY AND THEN SOLVE
FOR DY/DX. SO ON THE LEFT SIDE WE WOULD HAVE 1/Y x DY/DX.
ON THE RIGHT SIDE, WE HAVE 4 DIVIDED BY THE QUANTITY 2X - 3
+ THIS WOULD BE 30X DIVIDED BY THE QUANTITY 5X SQUARED + 2,
AND NOW TO SOLVE THIS FOR DY/DX WE'LL MULTIPLY BOTH SIDES
OF THE EQUATION BY Y.
SO WE'D HAVE DY/DX IS EQUAL
TO Y x OUR TWO FRACTIONS 4 DIVIDED BY THE QUANTITY 2X - 3
+ 30X DIVIDED BY THE QUANTITY 5X SQUARED + 2.
EVEN THOUGH WE HAVE THIS SOLVED FOR DY/DX WE DON'T WANT
TO LEAVE IT IN THIS FORM BECAUSE WE DON'T WANT
TO LEAVE THIS Y IN OUR DERIVATIVE FUNCTION. WE CAN
REPLACE Y WITH THE ORIGINAL FUNCTION WHICH IS THE QUANTITY
2X - 3 SQUARED x THE QUANTITY 5X SQUARED + 2 CUBED. SO LET'S
GO AHEAD AND PERFORM THAT SUBSTITUTION ON THE NEXT SLIDE.
SO AGAIN WHAT WE'VE DONE HERE IS WE'VE SUBSTITUTED THE
ORIGINAL FUNCTION FOR THIS Y HERE. BUT NOW WE'LL
GO AND FIND THIS PRODUCT. WHEN MULTIPLYING BY THIS FIRST
FRACTION NOTICE HOW ONE FACTOR OF 2X - 3 IS GOING TO SIMPLIFY
OUT SO THIS FIRST PRODUCT WOULD BE 4 x ONE FACTOR OF 2X - 3 x
3 FACTORS OF 5X SQUARED + 2 AND THEN WHEN MULTIPLYING BY THE
SECOND FRACTION NOTICE HOW ONE FACTOR OF 5X SQUARED + 2 WOULD
SIMPLIFY OUT. SO THAT WOULD LEAVE US
WITH 30X AND THEN TWO FACTORS OF THE QUANTITY 2X - 3 AND TWO
FACTORS OF THE QUANTITY 5X SQUARED + 2.
NOW WE'RE GOING TO GO AHEAD AND FACTOR THIS BY FACTORING OUT
THE GREATEST COMMON FACTOR
OF THESE TWO PRODUCTS. SO THE GREATEST COMMON FACTOR WOULD BE
2 AND THEN ONE FACTOR OF 2X - 3 AND TWO FACTORS OF 5X SQUARED
+2. IF WE FACTOR THIS OUT OF THE FIRST PRODUCT,
IT WOULD LEAVE US WITH A FACTOR OF 2 AND THEN ONE
FACTOR OF 5X SQUARED + 2. IF WE FACTOR THIS OUT
OF THE SECOND PRODUCT, NOTICE HOW IT WOULD LEAVE US
WITH THE FACTOR OF 15X AND ONE FACTOR
OF THE QUANTITY 2X - 3. SO AGAIN, YOU CAN SEE A LOT OF
THIS PROBLEM IS JUST ALGEBRA. NOW WE'RE GOING
TO SIMPLIFY THIS INSIDE THE BRACKETS. SO WE'LL DISTRIBUTE
THE 2 HERE AND DISTRIBUTE THE 15X HERE. SO WE'LL HAVE DY/DX =
2 x THE QUANTITY 2X - 3 x THE QUANTITY 5X SQUARED + 2
SQUARED. THEN WE'LL HAVE 10X TO THE 2nd + 4 AND THEN + 30X
TO THE 2nd AND THEN - 45X AND NOW WE'LL GO AHEAD AND COMBINE
THE LIKE TERMS HERE. SO WE FINALLY HAVE DY/DX = 2 x THE
QUANTITY 2X - 3 x THE QUANTITY 5X SQUARED + 2 SQUARED AND THEN
HERE WE HAVE 10X SQUARED + 30X SQUARED THAT'S 40X SQUARED -
45X + 4. SO THIS WOULD BE OUR DERIVATIVE FUNCTION USING
LOGARITHMIC DIFFERENTIATION.
I HOPE YOU FOUND THIS EXPLANATION HELPFUL.