[ MUSIC ]
- WELCOME TO A VIDEO 
ON THE DERIVATIVES
OF LOGARITHMIC FUNCTIONS.
LET'S GO AHEAD AND GET STARTED.
HERE ARE OUR RULES.
THE DERIVATIVE OF NATURAL LOG 
X IS EQUAL TO 1/X.
AND THE SECOND FORMULA HERE 
WE HAVE A COMPOSITE FUNCTION.
WE HAVE THE DERIVATIVE 
EQUAL TO 1/U x U PRIME.
IF WE HAVE THE DERIVATIVE 
OF A LOGARITHM
OTHER THAN BASE E WE'RE GOING 
TO HAVE THE DERIVATIVE
OF LOG "A" 
BASE X = 1/NATURAL LOG "A" x X.
NOTICE "A" IS OUR BASE.
AND IF WE HAVE A COMPOSITE 
FUNCTION INVOLVING A LOGARITHM
HERE'S OUR DERIVATIVE FORMULA.
I'LL BRIEFLY OUTLINE THE PROOF 
OF THE DERIVATIVE
OF NATURAL LOG X.
ESSENTIALLY WHAT WE DO 
IS REWRITE THIS
AS AN EXPONENTIAL EQUATION,
TAKE THE DERIVATIVE 
OF BOTH SIDES,
IMPLY IMPLICIT DIFFERENTIATION, 
SOLVE FOR DYDX TO OBTAIN
OUR DERIVATIVE FORMULA.
FOR THE DERIVATIVE 
OF LOG BASE "A" OF X,
WE WOULD APPLY THE CHANGE 
OF BASE FORMULA,
THEN TAKE THE DERIVATIVE 
OF THIS, WHICH WOULD GIVE US
1/NATURAL LOG "A" x 1/X 
WITH RESULTS
IN OUR DERIVATIVE FORMULA.
AND THEN OF COURSE IN BOTH CASES 
WE WOULD APPLY THE CHAIN RULE
WHEN NEEDED.
LET'S GO AHEAD AND PRACTICE.
WE WANT TO FIND THE DERIVATIVE 
OF THIS FUNCTION.
FIRST THING WE RECOGNIZE 
IS THIS IS A COMPOSITE FUNCTION,
SO THERE'S OUR U, 6X SQUARED.
SO OUR DERIVATIVE WOULD EQUAL 
1/U x U PRIME.
1/6X SQUARED x 12X, 
WHICH SIMPLIFIES TO 2/X.
WE COULD'VE BROKEN THIS DOWN 
INTO A SUM OF TWO LOGS
USING THE PROPERTIES OF LOGS.
AND FOR THIS ONE 
I GUESS I'LL SHOW IT.
THIS WOULD 
BE THE SAME
AS NATURAL LOG 6 + 
NATURAL LOG X SQUARED.
THEN, IF WE WANTED TO, 
WE COULD APPLY
THE POWER PROPERTY AND MOVE 
THIS 2 TO THE POSITION
OF THE COEFFICIENT.
SO WE'D HAVE NATURAL LOG 6 
+ 2 NATURAL LOG X,
THEREFORE OUR DERIVATIVE 
FUNCTION WOULD BE EQUAL
TO THE DERIVATIVE 
OF THESE TWO TERMS,
WHICH WOULD BE 0 + 2 x 1/X 
OR 2/X.
SO OFTEN THERE'S MORE THAN ONE 
WAY TO FIND THE DERIVATIVE
OF A LOG FUNCTION.
BUT IN MANY CASES 
I'LL JUST LEAVE IT
IN EXISTING FORM AND APPLY 
THE APPROPRIATE
DERIVATIVE FORMULA.
FOR NUMBER TWO, NOTICE WE HAVE 
A QUOTIENT OF TWO FUNCTIONS,
SO WE ARE GOING TO HAVE TO APPLY 
THE QUOTIENT RULE.
FOR THE SAKE OF TIME 
I'VE ALREADY SET UP
THE QUOTIENT RULE.
YOU MAY WANT TO PAUSE 
TO VERIFY THIS.
LET'S GO AHEAD 
AND DO THE NEXT STEP
BY FINDING THESE DERIVATIVES.
WE HAVE X TO THE 4TH x 1/U 
x U PRIME OR JUST U PRIME/1
THAT WOULD BE 2/2X - THE NATURAL 
LOG OF 2X x 4X CUBED/X
TO THE 8TH.
LET'S SIMPLIFY HERE. 
THE TWOS SIMPLIFY.
THIS X SIMPLIFIES 
WITH ONE OF THOSE FACTORS OF X.
SO WE'D HAVE 
X TO THE 3RD - 4X CUBE
NATURAL LOG 2X/X TO THE 8TH.
NOTICE THAT ALL THESE TERMS 
DO HAVE A COMMON FACTOR
OF X CUBED.
SO IF I FACTOR OUT X CUBE 
FROM THE NUMERATOR
AND REWRITE THE DENOMINATOR, 
WE CAN SEE THAT X CUBED/X CUBED
WOULD GIVE US 1.
SO OUR FINAL DERIVATIVE 
SIMPLIFIES TO THIS.
NEXT, THE FIRST THING WE SHOULD 
RECOGNIZE IS WE HAVE A PRODUCT
OF TWO FUNCTIONS, 
SO WE ARE GOING TO HAVE TO APPLY
THE PRODUCT RULE.
THE ONE THING THAT WE COULD DO 
IN THIS CASE IS THAT WE COULD
TAKE THIS EXPONENT OF 2 
AND MOVE IT TO THE FRONT
OF THE NATURAL LOG.
IN THIS CASE I WILL DO IT.
SO OUR FUNCTION 
WOULD ACTUALLY BE
E TO THE POWER OF 2X x 2 
NATURAL LOG X.
SO WHEN WE APPLY 
OUR PRODUCT RULE
THIS WILL BE OUR FUNCTION F, 
OUR FIRST FUNCTION.
AND THIS WILL BE OUR G, 
OUR SECOND FUNCTION.
AND I'VE ALREAD SET UP
THE PRODUCT RULE AGAIN
JUST TO SAVE SOME TIME.
FIRST (INAUDIBLE) 2ND + THE 
2ND X THE DERIVATIVE OF THE 1ST.
LET'S GO AHEAD 
AND FIND OUR DERIVATIVES.
THE DERIVATIVE 
OF 2 NATURAL LOG X
WOULD BE 2 x 1/X + 2 
NATURAL LOG X x THE DERIVATIVE
OF 2 E TO THE X.
REMEMBER THE DERIVATIVE 
OF E TO THE U
IS E TO THE U x U PRIME.
SO WE'D HAVE E TO THE 2X--SO 
WE'D HAVE E TO THE 2X x 2,
WHICH WOULD SIMPLIFY TO THIS. 
NOTICE WE HAVE TWO FRACTIONS.
SO IF WE WANTED TO COMBINE 
THESE FRACTIONS
WE'D HAVE TO OBTAIN 
A COMMON DENOMINATOR
AND MULTIPLY THIS FRACTION 
BY X OVER X,
WHICH WILL RESULT IN THIS.
NOTICE JUST THE EXTRA FACTOR 
OF X IN THIS TERM.
NOW, THIS NUMERATOR DOES FACTOR, 
WHICH WOULD BE OPTIONAL
BECAUSE IT'S NOT GOING 
TO SIMPLIFY.
BUT I'LL GO AHEAD 
AND FACTOR OUT
THE COMMON FACTOR OF 2 E 
TO THE 2X,
WHICH WILL RESULT IN 1 + 2X 
NATURAL LOG X
INSIDE THE PARENTHESIS.
AGAIN, ON THIS PROBLEM I
DO HAVE THE OPTION OF REWRITING
THIS AS LOG X + LOG OF X - 9.
BUT I'LL LEAVE IT 
IN ITS CURRENT FORM
AND JUST USE A U VALUE 
OF X SQUARED - 9X.
HERE'S OUR DERIVATIVE FORMULA 
FOR LOG BASE "A" OF U.
SO APPLYING OUR DERIVATIVE 
FORMULA, OUR DENOMINATOR'S
GOING TO BE NATURAL LOG "A" x U, 
SO THE NATURAL LOG
OF 4 x X SQUARED - 9X x U PRIME.
DERIVATIVE OF OUR U VALUE 
WOULD BE 2X - 9.
NOTHING'S GOING 
TO SIMPLIFY HERE,
SO WE HAVE 2X - 9/NATURAL LOG 
4 x X SQUARED - 9X.
OKAY. 
LET'S TRY ANOTHER.
NOW, IF I LEAVE THIS 
IN ITS CURRENT FORM,
MY U VALUE IS GOING TO BE 
A QUOTIENT
AND I DON'T WANT THAT.
SO I'M GOING TO GO AHEAD 
AND REWRITE THIS
AS H OF X IS EQUAL 
TO NATURAL LOG
OF X SQUARE - 5 
- NATURAL LOG OF X.
AND I'LL FIND THE DERIVATIVE 
OF THE FUNCTION IN THIS FORM.
SO HERE WE HAVE A U VALUE 
OF X SQUARED - 5,
SO OUR DERIVATIVE HERE 
WILL BE 1/U x U PRIME.
1/X SQUARED - 5 x U PRIME, 
WHICH IS 2X, - DERIVATIVE
OF NATURAL LOG X OR JUST 1/X.
NOW, I DO WANT TO COMBINE 
THESE FRACTIONS
SO I'LL OBTAIN 
MY COMMON DENOMINATOR.
SO THE RESULT HERE 
WILL BE 2X SQUARED - X SQUARED
+ 5 OVER OUR COMMON DENOMINATOR, 
WHICH SIMPLIFIES
TO X SQUARED + 5/X 
x X SQUARED - 5.
AND THAT WOULD BE 
OUR DERIVATIVE FUNCTION.
LET'S TAKE A LOOK AT ONE MORE.
OKAY. ON THIS PROBLEM WE WANT 
TO FIND THE EQUATION
OF THE TANGENT LINE 
TO THE FUNCTION
AT THE GIVEN POINT.
ONE THING I WANT TO POINT OUT 
HERE IS THAT HERE WE HAVE
THE ENTIRE LOGARITHM 
BEING SQUARED,
SO WE'RE NOT ALLOWED TO TAKE 
THIS 2 AND MOVE IT TO THE FRONT.
IF IT WAS JUST THE X 
BEING SQUARED WE COULD APPLY
THE POWER PROPERTY, 
BUT NOT IN THIS FORM.
SO IN ORDER TO FIND THE SLOPE 
WITH THE TANGENT LINE,
OF COURSE, WE HAVE TO FIND 
THE DERIVATIVE.
SO OUR U VALUE IS GOING 
TO BE NATURAL LOG X.
SO WE'RE GOING TO TREAT THIS 
AS IF IT WAS U SQUARED
AND APPLY THE POWER RULE.
SO F PRIME OF X IS EQUAL 
TO 2U TO THE 1ST x U PRIME
OR 2 NATURAL LOG X 
TO THE 1ST POWER x U PRIME,
WHICH IS 1/X.
SO OUR DERIVATIVE FUNCTION 
WILL BE 2 NATURAL LOG X
DIVIDED BY X.
AND, NO, THESE X's 
DO NOT SIMPLIFY.
SO IF WE WANT TO FIND THE SLOPE 
WITH THE TANGENT LINE
WE HAVE TO EVALUATE 
THE DERIVATIVE FUNCTION
AT X = 0.5.
THIS IS APPROXIMATELY - 2.7726.
NOW, TO SAVE SOME TIME 
I'VE ALREADY DETERMINED
THAT THE APPROXIMATE Y VALUE 
OF THIS POINT IS 0.4805.
SO THE EQUATION OF OUR TANGENT 
LINE WILL BE Y - Y1,
0.4805 = -2.7726 x X - 0.5.
IF WE SOLVE THIS FOR Y 
WE WILL OBTAIN THE EQUATION
Y = -2.7726X + 1.8668.
NOW, THERE IS A QUICK WAY 
TO DETERMINE IF WE FOUND
THE VALUE 
OF THIS DERIVATIVE CORRECTLY.
IF YOU GO TO YOUR GRAPHING 
CALCULATOR AND HIT MATH,
OPTION 8, THEN YOU TYPE 
IN THE FUNCTION, COMMA, X,
COMMA, THE DESIRED VALUE OF X, 
0.5, AND HIT ENTER.
IT WILL FIND THE VALUE 
OF THE DERIVATIVE
AT A GIVEN X VALUE.
SO WE DID OUR WORK CORRECTLY.
LET'S GO AHEAD 
AND VERIFY THIS
WITH A GRAPH.
NOTICE WE HAVE A Y INTERCEPT 
THAT'S CLOSE TO 2
AND A SLOPE THAT'S CLOSE TO -3.
HERE'S OUR POINT OF TANGENCY.
AND YOU CAN SEE WE HAVE 
Y INTERCEPT THAT'S ROUGHLY 2,
AND A SLOPE THAT'S ROUGHLY -3.
OUR WORK HAS BEEN VERIFIED.
THANK YOU FOR WATCHING.
