- WELCOME TO A VIDEO 
ON DERIVATIVE
OF AN EXPONENTIAL FUNCTION.
THE GOAL OF THIS VIDEO
IS TO DIFFERENTIATE 
EXPONENTIAL FUNCTIONS.
LET'S GO AHEAD AND TAKE A LOOK 
AT THE RULES.
IF OUR BASE IS E,
THE DERIVATIVE OF E TO THE X 
IS EQUAL TO E TO THE X.
THIS IS ACTUALLY QUITE AMAZING.
THIS FUNCTION 
IS ITS OWN DERIVATIVE,
MEANING IF WE EVALUATE THE 
FUNCTION AT A GIVEN VALUE OF X,
THE FUNCTION VALUE ALSO TELLS US 
THE SLOPE OF THE TANGENT LINE
AT THAT X VALUE.
THE SECOND FORMULA HERE 
JUST APPLIES THE CHAIN RULE.
IF THE EXPONENT IS SOME FUNCTION 
IN TERMS OF X,
WE'D APPLY THE CHAIN RULE.
AND THE DERIVATIVE WOULD BE E 
TO THE U x A U PRIME.
NOW, IF THE BASE 
OF THE EXPONENTIAL IS NOT E,
THEN NOTICE THE DERIVATIVE 
WOULD HAVE AN EXTRA FACTOR
OF NATURAL LOG "A" x "A" 
TO THE X.
AND THEN, AGAIN, 
THIS LAST RULE IS
IF WE HAVE A RAISED 
TO SOME FUNCTION IN TERMS OF X,
WE WOULD HAVE TO APPLY 
THE CHAIN RULE.
SO WE'D HAVE NATURAL LOG "A," 
"A" TO THE U x U PRIME.
LET'S TAKE A QUICK LOOK AT 
A ROUGH PROOF OF THE DERIVATIVE
OF THE FUNCTION F OF X 
IS EQUAL TO E TO THE X.
WE START WITH THE BASIC 
EXPONENTIAL FUNCTION,
APPLY THE LIMIT DEFINITION 
OF DERIVATIVE,
DO OUR SUBSTITUTION, 
DO SOME FANCY ALGEBRA HERE,
FACTOR OUT "A" TO THE X.
AND WHAT WE HAVE IS "A" TO THE X 
x THIS SPECIAL LIMIT.
AND WHAT WE'RE GOING 
TO ARGUE NOW
IS IF WE KNEW THE VALUE OF "A"
THAT WOULD MAKE THIS LIMIT 
EQUAL TO 1,
THEN WE WOULD HAVE THE CASE 
WHERE THE EXPONENTIAL FUNCTION
IS ITS OWN DERIVATIVE.
AND SO WHAT WE DO IS WE TAKE 
THAT SPECIAL CASE OVER HERE.
WHAT WOULD "A" HAVE TO BE IN 
ORDER FOR THIS LIMIT TO EQUAL 1?
WELL, IF WE ASSUME ITS EQUAL 
TO 1, SOLVE IT FOR "A" TO THE H,
AND RAISE BOTH SIDES OF THE 
EQUATION TO THE POWER OF 1/H,
WE'D HAVE "A" 
EQUAL TO THIS EXPRESSION HERE.
WHICH, WE COULD THEN CONCLUDE 
THAT THIS LIMIT IS EQUAL TO 1
THEN "A" WOULD BE EQUAL 
TO THIS LIMIT HERE.
AND BY DEFINITION, 
THIS LIMIT IS EQUAL TO E,
THEREFORE, WE HAVE "A" 
EQUAL TO E.
SO WHEN "A" IS EQUAL TO E, GOING 
BACK OVER TO THE LEFT SIDE,
WE HAVE THE DERIVATIVE OF E 
TO THE X IS EQUAL TO E TO THE X.
NOW, YOU MAY WANT TO PAUSE
AND TAKE A CLOSER LOOK 
AT THIS PROOF,
BUT THAT'S THE GENERAL IDEA.
NOW, THE NEXT CASE IS WHAT WE DO 
IF THE BASE ISN'T E.
WELL, FROM SOME OF PROPERTIES 
WE CAN REWRITE "A" TO THE X
AS E TO THE POWER OF X 
x NATURAL LOG "A".
BUT IN THIS FORM WE CAN 
JUST APPLY THE CHAIN RULE
TO THE DERIVATIVE OF E TO THE U.
SO WHAT WE'RE SAYING HERE IS 
THAT DERIVATIVE OF "A" TO THE X
IS EQUAL TO THIS DERIVATIVE.
SO IF WE LET THIS EQUAL E 
TO THE U,
THE DERIVATIVE OF THIS FUNCTION 
WOULD BE E TO THE U x U PRIME.
NOW, NATURAL LOG "A" 
IS JUST THE CONSTANT.
THE DERIVATIVE OF U 
IS JUST 1 NATURAL LOG "A,"
WHICH GIVES US NATURAL LOG 
"A" x "A" TO THE X
WHEN "A" DOES NOT EQUAL E.
AGAIN, YOU MAY WANT 
TO PAUSE THE VIDEO,
TAKE A CLOSER LOOK AT THIS, 
BUT THAT'S THE GENERAL IDEA.
LET'S GO AHEAD 
AND PUT IT INTO PRACTICE.
LET'S SAY WE WANT TO FIND 
THE DERIVATIVE
OF EACH OF THESE FUNCTIONS.
SO FOR THE DERIVATIVE 
OF THIS FUNCTION--
SO TO FIND THE DERIVATIVE 
OF THIS FUNCTION
WE ARE GOING TO NEED 
THIS DERIVATIVE FORMULA
E TO THE U x U PRIME
WHERE OUR VALUE OF U 
IS EQUAL TO 6X.
SO FIRST, WE HAVE THE DERIVATIVE 
OF THIS TERM,
WHICH WOULD BE 10X 
TO THE FOURTH - 3 x--
HERE'S OUR FORMULA, 
E TO THE U x U PRIME.
THAT WOULD BE E 
TO THE POWER OF 6X x U PRIME.
THE DERIVATIVE OF 6X WOULD BE 6.
WE CAN CLEAN THIS UP TO 10X 
TO THE FOURTH - 18 E
TO THE POWER OF 6X.
NOW, ON NUMBER TWO 
WE HAVE TO REALIZE
THAT THIS IS A PRODUCT 
OF TWO FUNCTIONS.
WE'LL CALL THE FIRST FUNCTION F 
AND THE SECOND FUNCTION G.
SO WE'RE GOING TO HAVE 
TO APPLY THE PRODUCT RULE.
AND SO WHAT I'VE DONE ALREADY 
IS SET THIS UP.
WE HAVE THE FIRST FUNCTION TIMES 
THE DERIVATIVE OF THE SECOND,
PLUS THE SECOND TIMES 
THE DERIVATIVE OF THE FIRST.
LET'S GO AHEAD AND FIND THESE 
DERIVATIVES AND THEN SIMPLIFY.
NOW, THE DERIVATIVE OF E 
TO THE POWER OF -2X,
WE ARE GOING TO HAVE TO APPLY 
THE CHAIN RULE AGAIN.
SO OUR DERIVATIVE 
IS E TO THE U x U PRIME.
SO WE'D HAVE E 
TO THE POWER OF -2X x -2
+ E TO THE -2X x THE DERIVATIVE 
OF X TO THE THIRD
THAT WOULD BE 3X SQUARED.
WE HAVE -2X CUBED, E TO THE -2X 
+ 3X SQUARED, E TO THE -2X.
NOW, I WILL FACTOR THIS ONE.
WE CAN SEE THESE TWO TERMS HAVE 
A COMMON FACTOR OF X SQUARED E
TO THE -2X.
NOW, IF WE FACTORED THIS OUT 
OF THE FIRST TERM
WE'D BE LEFT WITH -2X.
IF WE FACTORED X SQUARED E 
TO THE -2X OUT OF THIS TERM
WE'D HAVE A +3.
SO I'LL PUT THE +3 FIRST 
AND THE -2X SECOND.
AND NOW THERE'S A DERIVATIVE 
IN FACTORED FORM.
LET'S GO AHEAD AND TRY ANOTHER.
HERE WE HAVE A QUOTIENT,
SO ON THE RIGHT SIDE HERE 
I WROTE OUT THE QUOTIENT RULE.
REMEMBER ITS LOW D HIGH 
MINUS HIGH D LOW, LOW, LOW.
AND OVER HERE ON THE LEFT 
I'VE ALREADY SET IT UP.
WE HAVE THE DENOMINATOR TIMES 
DERIVATIVE OF THE NUMERATOR,
MINUS THE NUMERATOR, TIMES 
THE DERIVATIVE OF DENOMINATOR
ALL OVER THE DENOMINATOR 
SQUARED.
THE DERIVATIVE OF E 
TO THE POWER OF 3X
WOULD BE E TO THE 3X 
x U PRIME -- 3 - E TO THE 3X
x 6X TO THE 5th/X TO THE 12th.
LET'S CLEAN THIS UP AND THEN SEE 
IF WE CAN FACTOR IT.
NOW, I NOTICED THE NUMERATOR 
HAS A COMMON FACTOR OF 3X
TO THE 5th,
E TO THE 3X.
LET'S FACTOR THAT OUT.
SO WE'D BE LEFT WITH X - 2/X 
TO THE 12th.
AND WE CAN SEE THAT WE HAVE X TO 
5th HERE AND X TO THE 12th HERE.
THIS WOULD SIMPLIFY AND THIS 
WOULD BECOME X TO THE 7th.
SO OUR FINAL DERIVATIVE 
WOULD LOOK LIKE THIS.
OKAY. LET'S TRY 
ANOTHER QUESTION.
DETERMINE THE SLOPE 
OF A TANGENT LINE
TO THE FUNCTION 
AT THE GIVEN POINT.
WELL, WE KNOW IN ORDER TO FIND 
THE SLOPE OF THE TANGENT LINE
WE HAVE TO FIND THE DERIVATIVE 
OF THIS FUNCTION.
LET'S GO AHEAD AND DO THAT.
IT WOULD BE 2 x THE DERIVATIVE 
OF E TO THE POWER OF 3X.
AGAIN, HOPEFULLY WE'RE 
PRETTY GOOD NOW RECOGNIZING
THAT OUR U WOULD EQUAL -3X.
SO WE'RE GOING TO HAVE E 
TO THE U, E TO THE -3X x -3.
SO OUR DERIVATIVE FUNCTION 
IS -6, E TO THE -3X.
WE WANT TO FIND THE SLOPE WITH 
A TANGENT LINE AT THIS POINT.
WE NEED TO EVALUATE 
THIS FUNCTION AT X = 0.
OF COURSE E TO THE 0 WOULD BE 1.
-6 x 1 GIVES US A SLOPE OF -6.
LET'S GO AHEAD AND GRAPH THIS 
FUNCTION AND CHECK OUR WORK.
SO HERE'S OUR FUNCTION, 
HERE'S OUR POINT OF TANGENCY.
THE QUESTION IS, DOES THIS LINE 
HAVE A SLOPE OF -6.
WE START UP HERE AND GO DOWN 
SIX UNITS AND RIGHT ONE,
WE HAVE THE CORRECT LINE, 
WHICH VERIFIES OUR WORK.
LET'S GO AHEAD AND TAKE A LOOK
AT ONE MORE APPLICATION 
QUESTION.
MORE AMERICANS ARE BUYING 
ORGANIC FOOD AND VEGETABLES
MADE WITH ORGANIC INGREDIENTS.
THE AMOUNT "A" OF T IN BILLIONS 
OF DOLLARS SPENT ON ORGANIC FOOD
T YEARS AFTER 1995 CAN BE 
APPROXIMATED BY THIS FUNCTION.
WE HAVE TWO QUESTIONS 
TO CONSIDER.
ESTIMATE THE AMOUNTS 
THAT AMERICANS WILL SPEND
ON ORGANIC FOOD 
IN THE YEAR 2009.
SO THE QUESTION IS WHAT VALUE 
OF T ARE WE GOING TO USE.
WELL, BASED UPON THE QUESTION, 
1995 WAS OUR BASE YEAR.
SO WE HAVE TO TAKE OUR DESIRED 
YEAR AND SUBTRACT 1995.
THIS WILL GIVE US 14.
SO THIS IS NOT A CALCULUS 
QUESTION.
WE JUST NEED TO EVALUATE "A" 
AT A VALUE OF 14 FOR T.
SECOND NATURAL LOG 
BRINGS UP THE BASE E,
AND WE HAVE APPROXIMATELY 30.2.
NOW, IF YOU RECALL FROM THE 
QUESTION, IT WAS $30.2 BILLION,
"A" OF T WAS IN BILLIONS.
AND THE LAST QUESTION,
ESTIMATE THE RATE AT WHICH 
THE SPENDING ON ORGANIC FOOD
WAS GROWING IN THE YEAR 2006.
WHENEVER WE HEAR THAT WORD 
"RATE" OR "RATE OF CHANGE"
WE SHOULD BE THINKING 
DERIVATIVE.
SO WE NEED TO EVALUATE 
THE DERIVATIVE.
LET'S START BY FINDING OUR VALUE 
OF T.
THAT'S GOING TO GIVE US A 
T VALUE OF 11 FOR THIS QUESTION.
BUT ON THIS ONE WE WANT TO FIND 
THE RATE OF CHANGE.
WE NEED TO FIND THE DERIVATIVE.
REMEMBER THE DERIVATIVE OF THIS 
WOULD BE E TO THE U x U PRIME.
OF COURSE T PRIME--OF COURSE 
U PRIME WOULD BE 0.18.
SO SIMPLIFYING THIS,
WE WOULD OBTAIN 
THE DERIVATIVE FUNCTION HERE.
SO IF WE SIMPLIFY THIS WE'D 
OBTAIN THIS DERIVATIVE FUNCTION.
AND NOW WE NEED TO EVALUATE THIS 
AT T = 11.
WHICH, GIVES US APPROXIMATELY 
3.17.
AGAIN, THIS IS IN BILLIONS 
OF DOLLARS, BUT IT'S A RATE.
SO BILLIONS OF DOLLARS, 
AND TIME IS IN YEAR,
SO PER YEAR AT THAT INSTANT 
MORTISE THE 11th.
NOW I WANT TO GO BACK 
TO THE GRAPHING CALCULATOR
JUST FOR A MOMENT
AND SHOW YOU ANOTHER WAY 
YOU CAN CHECK YOUR WORK.
THIS GRAPHING CALCULATOR 
WILL EVALUATE DERIVATIVES
AT A SPECIFIC VALUE OF X, 
OR IN THIS CASE T.
SO IF WE HIT OUR MATH KEY, 
AND SCROLL DOWN TO OPTION 8.
IF WE TYPE IN 
THE ORIGINAL FUNCTION,
AND I'M GOING TO USE X 
INSTEAD T, THEN TYPE IN COMMA X.
COMMA THE VALUE OF X WE WANT 
TO DETERMINE THE DERIVATIVE AT,
WHICH IS 11.
IT WILL GIVE US THE SAME ANSWER.
IT'S A NICE WAY TO CHECK,
BUT I'M SURE YOUR INSTRUCTOR 
WILL WANT YOU TO SHOW
THE DERIVATIVE FUNCTION.
OKAY, I HOPE THAT HELPS EXPLAIN
HOW TO FIND THE DERIVATIVE 
OF EXPONENTIAL FUNCTIONS.
THANK YOU FOR WATCHING.
