- WELCOME BACK. HOPEFULLY YOU JUST WATCHED THE VIDEO
WHERE WE FOUND THE DERIVATIVE OF THE GIVEN FUNCTION
WITHOUT USING THE QUOTIENT RULE.
SINCE OUR DENOMINATOR IS A SINGLE TERM
WE WERE ABLE TO DIVIDE EACH TERM
IN THE NUMERATOR BY X TO THE 1/2,
SIMPLIFY AND THEN APPLY THE BASIC POWER RULE
INSTEAD OF THE QUOTIENT RULE.
BUT IN THIS VIDEO I WANT TO SHOW HOW IF WE DO APPLY
THE QUOTIENT RULE WHILE IT IS GOING TO BE MORE WORK
WE WILL OBTAIN THE SAME DERIVATIVE FUNCTION.
SO FOR REFERENCE HERE'S THE QUOTIENT RULE
WHERE THE DERIVATIVE OF F DIVIDED BY G
WITH RESPECT TO X IS EQUAL TO G x F PRIME - F x G PRIME
DIVIDED BY G SQUARED.
SO FOR OUR FUNCTION, F IS EQUAL TO X SQUARED + 4X - 3
AND G IS EQUAL TO THE SQUARE ROOT OF X.
BUT WE'RE ACTUALLY GOING TO WRITE THE SQUARE ROOT OF X
USING RATIONAL EXPONENTS.
REMEMBER SQUARE ROOT X IS EQUAL TO X TO THE 1/2
BECAUSE THE EXPONENT ON X IS 1 AND THE INDEX IS 2.
SO OUR DERIVATIVE FUNCTION WILL HAVE THE DENOMINATOR
OF SQUARE ROOT X SQUARED OR X TO THE 1/2 SQUARED
AND THEN FOR THE NUMERATOR WILL HAVE THE DENOMINATOR X
TO THE 1/2 x THE DERIVATIVE
OF THE NUMERATOR - THE NUMERATOR x THE DERIVATIVE
OF THE DENOMINATOR.
SO AGAIN, WE HAVE (G x F PRIME) - (F x G PRIME)
DIVIDED BY G SQUARED.
NOTICE HOW WE HAVEN'T FOUND ANY DERIVATIVES YET
WE JUST WROTE OUT THE QUOTIENT RULE.
SO NOW WE'LL GO BACK AND FIND THE DERIVATIVE HERE, HERE,
AND THEN BEGIN TO SIMPLIFY.
NOTICE IN THE DENOMINATOR WE HAVE X TO THE 1/2 SQUARED
SO WE'D MULTIPLY THE EXPONENTS 1/2 x 2 IS 1.
SO WE HAVE X TO THE 1st OR JUST X IN THE DENOMINATOR
AND THEN WE HAVE (X TO THE 1/2 x THE DERIVATIVE OF X SQUARED)
+ (4X - 3).
THE DERIVATIVE OF X SQUARED IS 2X.
THE DERIVATIVE OF 4X IS 4.
THE DERIVATIVE OF 3 IS ZERO.
SO THE DERIVATIVE IS (2X + 4) - THE QUANTITY X SQUARED
+ (4X - 3) x THE DERIVATIVE OF X TO THE 1/2.
SO APPLYING THE POWER RULE WE'LL HAVE 1/2 x X
TO THE POWER OF 1/2 - 1 IS -1/2.
SO NOW WE'LL DISTRIBUTE HERE AS WELL AS HERE 1, 2, 3.
SO F PRIME OF X CURRENTLY HAS A DENOMINATOR OF X.
HERE WE'D HAVE X TO THE 1/2 x 2X TO THE 1st POWER.
SO WE'RE GOING TO ADD THE EXPONENTS.
WE'LL HAVE 2X TO THE 3/2 + 4X TO THE 1/2 - 1/2
X TO THE POWER OF -1/2 + 2 THAT'S 3/2.
4 x 1/2 IS 2 SO + 2 X TO THE POWER OF -1/2 + 1
THAT'S TO THE 1/2
AND THEN FINALLY WE HAVE -3/2 X TO THE NEGATIVE 1/2.
NOW WE'RE GOING TO CLEAR THE PARENTHESES
AND COMBINE LIKE TERMS.
TO CLEAR THE PARENTHESES WE CAN THINK OF DISTRIBUTING A 1 HERE
BUT BECAUSE OF THE MINUS HERE WE WOULD DISTRIBUTE -1
TO CLEAR THE SECOND SET OF PARENTHESES.
LET'S GO AHEAD AND DO THAT ON THE NEXT SLIDE.
OKAY, SO MULTIPLYING BY 1 IS NOT GOING TO CHANGE ANYTHING
SO WE HAVE 2X TO THE 3/2 + 4X TO THE 1/2.
MULTIPLYING BY -1 HERE IS GOING TO CHANGE THE SIGN OF EACH TERM
SO WE'LL HAVE - 1/2 X TO THE 3/2 - 2 X TO THE 1/2
AND THEN + 3/2 X TO THE -1/2.
NOW WE'LL GO AHEAD AND COMBINE LIKE TERMS
MEANING THE VARIABLE WILL BE THE SAME AS WELL AS THE EXPONENT.
SO THESE TWO TERMS ARE LIKE TERMS
AND THESE TWO TERMS ARE LIKE TERMS.
SO WE HAVE F PRIME OF X = 2X TO THE 3/2 - 1/2X TO THE 3/2.
2 - 1/2 IS 3/2 SO WE HAVE 3/2 X TO THE 3/2
AND THEN 4X TO THE 1/2 - 2X TO THE 1/2 IS + 2X TO THE 1/2
AND THEN FINALLY WE STILL HAVE + 3/2 X TO THE -1/2.
ALL THIS IS STILL DIVIDED BY X BUT NOW AGAIN;
BECAUSE WE'RE DIVIDING BY A SINGLE TERM OR A MONOMIAL
WE CAN DIVIDE EACH TERM IN THE NUMERATOR BY X TO SIMPLIFY THIS.
SO WE'D HAVE F PRIME OF X = 3/2.
X TO THE 3/2 DIVIDED BY X WHICH IS ACTUALLY X
TO THE 1st + 2X TO THE 1/2 DIVIDED BY X TO THE 1st + 3/2
X TO THE -1/2 DIVIDED BY X TO THE 1st.
REMEMBER WHEN DIVIDING WE SUBTRACT THE EXPONENTS
IF THE BASES ARE THE SAME.
SO FINALLY WE HAVE F PRIME OF X = 3/2 X TO THE POWER OF--
3/2 - 1 IS 1/2 + 2 x X TO THE POWER OF 1/2 - 1 IS -1/2
AND HERE WE HAVE + 3/2 X TO THE POWER OF -1/2 - 1 IS -3/2.
LET'S CONTINUE ON THE NEXT SLIDE.
JUST AS BEFORE, WE COULD LEAVE THIS IN A VARIETY OF FORMS.
LET'S AT LEAST WRITE THIS SO THAT WE HAVE POSITIVE EXPONENTS
MEANING IF WE MOVE THESE TWO TO THE DENOMINATOR
IT'S GOING TO CHANGE THE SIGN OF THE EXPONENT.
SO WE'D HAVE F PRIME OF X = 3/2 X TO THE 1/2 + 2 DIVIDED BY X
TO THE 1/2 + 3 DIVIDED BY 2 X TO THE POWER OF 3/2.
SO THIS WOULD BE OUR DERIVATIVE FUNCTION
AND IF WE COMPARE THIS DERIVATIVE TO THE OTHER VIDEO
WHERE WE DIDN'T USE THE QUOTIENT RULE BECAUSE WE SIMPLIFIED
FIRST NOTICE HOW IT TOOK A LOT LESS WORK
BUT THE DERIVATIVE FUNCTION HERE IS STILL THE SAME.
NOW FOR THE LAST PART WE'RE ASKED TO FIND F PRIME OF 4
WHICH MEANS WE'LL JUST SUBSTITUTE 4 FOR X
INTO OUR DERIVATIVE FUNCTION
AND AGAIN, TO SAVE TIME I'VE ALREADY DONE THIS.
THIS SIMPLIFIES TO 67/16 WHICH IS EQUAL TO 4.1875
WHICH WILL BE THE SLOPE OF THE TANGENT LINE OF OUR FUNCTION
AT X = 4.
LET'S GO AHEAD AND LOOK AT THAT.
HERE'S OUR FUNCTION,
HERE'S THE POINT ON THE FUNCTION WHEN X = 4.
SO IF WE SKETCH A TANGENT LINE AT THAT POINT
IT MIGHT LOOK SOMETHING LIKE THIS AND SINCE F PRIME OF 4
WAS EQUAL TO 67/16
THIS WOULD BE THE SLOPE OF THIS BLUE TANGENT LINE.
I HOPE YOU FOUND THE COMPARISON OF THE TWO TECHNIQUES TO FIND
THE DERIVATIVE FUNCTION HELPFUL.