- WE WANT TO DETERMINE THE DERIVATIVE
OF THE GIVEN FUNCTION.
THE FIRST THING WE SHOULD NOTICE
IS WE HAVE A PRODUCT OF TWO FUNCTIONS.
SO IF WE CALL THE FIRST FUNCTION F AND THE SECOND FUNCTION G,
THE PRODUCT RULE TELLS US THE DERIVATIVE OF F x G
IS EQUAL TO F x G PRIME + G x F PRIME,
OR IF WE CONSIDER THE DERIVATIVE
IS EQUAL TO THE FIRST FUNCTION
x THE DERIVATIVE OF THE SECOND + THE SECOND FUNCTION
x THE DERIVATIVE OF THE FIRST.
THERE'S ONE MORE THING WE SHOULD NOTICE HERE.
IN ORDER TO FIND THE DERIVATIVE OF THE SQUARE ROOT OF X,
WE WANT TO REWRITE THIS AS A RATIONAL EXPONENT.
SO LET'S REWRITE THIS ONE MORE TIME.
F OF X IS EQUAL TO 2X TO THE 2nd - 5 x 3X + X TO THE 1/2 POWER.
SO WE'LL USE THIS FORM TO APPLY THE PRODUCT RULE.
SO NOW THAT ALL THE TERMS ARE IN THE CORRECT FORM,
WE CAN APPLY THE PRODUCT RULE.
SO THE DERIVATIVE IS GOING TO BE EQUAL TO F
OR OUR FIRST FUNCTION x THE DERIVATIVE OF G
WHICH IS OUR SECOND FUNCTION.
THE DERIVATIVE OF 3X IS 3
+ THE DERIVATIVE OF X TO THE 1/2 WOULD BE 1/2X TO THE 1/2 - 1
WHICH WILL BE -1/2 + THE SECOND FUNCTION
WHICH IS 3X + X TO THE 1/2 x THE DERIVATIVE OF THE FIRST
FUNCTION WHICH WILL JUST BE 4X.
NOW THAT WE HAVE FOUND THE DERIVATIVE, THE REST IS ALGEBRA.
WE HAVE TO MULTIPLY ALL THIS OUT AND THEN COMBINE LIKE TERMS.
SO WE'LL HAVE 4 PRODUCTS HERE, 1, 2, 3, 4,
AND WE'LL HAVE 2 PRODUCTS HERE WHEN WE DISTRIBUTE.
SO F PRIME OF X IS EQUAL TO,
WE'LL HAVE 6X SQUARED AND THEN 2X SQUARED x 1/2X TO THE -1/2.
WELL, 2 x 1/2 IS 1 AND X SQUARED x X TO THE -1/2,
WE ADD OUR EXPONENTS, WE'LL HAVE X TO THE 3/2.
SO + X TO THE 3/2, AND THEN -15
AND THEN WE'LL HAVE -5/2X TO THE -1/2
+ HERE WE'LL HAVE 12X SQUARED + 4X TO 3/2 POWER.
SO WE HAVE 2X SQUARED TERMS.
SO WE HAVE PRIME OF X IS GOING TO BE EQUAL TO,
WE'LL HAVE 18X SQUARED.
THERE ARE 2X TO THE 3/2 TERMS, 1 HERE AND 1 HERE.
REMEMBER YOU CAN THINK OF THIS AS 1X TO THE 3/2,
SO WE HAVE 5X TO THE 3/2.
THIS TERM HAS AN X TO THE -1/2 POWER,
SO LET'S GO AHEAD AND REWRITE THIS AS -5/2X TO THE 1/2.
REMEMBER I'M MOVING THIS TO THE DENOMINATOR,
WE CHANGE THE SINE OF THE EXPONENT, THEN WE HAVE -15.
AND THE LAST STEP YOU MIGHT WANT TO DO IS REWRITE X TO THE 1/2
AS THE SQUARE ROOT OF X.
SO WE HAVE 18X SQUARED + 5X TO THE 3/2 - 5/2
SQUARE ROOT X - 15, AND THEN WE HAVE OUR DERIVATIVE FUNCTION.
I GUESS I SHOULD MENTION ONE MORE THING.
X TO THE 3/2 COULD BE WRITTEN IN RADICAL FORM.
X TO THE 3/2 WOULD BE EQUAL TO THE SQUARE ROOT OF X
TO THE 3rd WHICH IS EQUAL TO 3 FACTORS OF X.
SO THERE IS A PERFECT SQUARE FACTOR HERE,
SO WOULD SIMPLIFY TO X SQUARE ROOT X.
SO WE COULD REWRITE X TO THE 3/2 AS X SQUARE ROOT X.
AND I BELIEVE SOME TEXTBOOKS MAY DO THAT,
ESPECIALLY IF THE GIVEN FUNCTION HAS A RADICAL IN IT.
I HOPE THIS EXAMPLE WAS HELPFUL.