- LET'S TAKE A LOOK AT A COUPLE MORE DERIVATIVES
THAT INVOLVE THE EXPONENTIAL FUNCTION WITH BASE E.
HERE WE HAVE F OF X = THE QUANTITY E
TO THE POWER OF 3X - 1 RAISED TO THE 5TH POWER.
SO WE'LL BE USING BOTH THE EXTENDED POWER RULE
AND THE DERIVATIVE FOR E TO THE U.
FIRST THING WE SHOULD NOTICE IS THAT WE HAVE
A COMPOSITE FUNCTION, THEREFORE, WE'LL LET
THE INNER FUNCTION EQUAL U.
SO U WILL BE E TO THE 3X - 1.
SO NOW WE CAN WRITE THIS AS U TO THE 5TH.
BUT WE KNOW AS SOON AS WE IDENTIFY U
WE'RE GOING TO NEED U PRIME.
SO NOW WE NEED TO DIFFERENTIATE E TO THE POWER OF 3X - 1.
BUT THIS IN ITSELF IS A COMPOSITE FUNCTION.
LOOKING AT OUR FORMULA HERE, FOR E TO THE 3X
WE'D HAVE TO LET U = 3X, BUT SINCE WE'VE ALREADY USED U
WE'LL GO AHEAD AND CALL THIS V.
SO THE DERIVATIVE OF E TO THE V WOULD BE E TO THE V x V PRIME.
SO U PRIME IS GOING TO BE EQUAL TO E TO THE 3X x THE DERIVATIVE
OF 3X, WHICH WOULD BE 3, AND THEN MINUS DERIVATIVE 1
WHICH IS 0.
SO THIS WOULD BE 3 E TO THE 3X.
SO NOW TO DETERMINE F PRIME OF X, WE CAN DIFFERENTIATE U
TO 5TH WITH RESPECTS TO X, WHICH WOULD BE 5 U
TO THE 4TH x U PRIME.
BUT U IS U TO THE POWER OF 3X - 1, AND U PRIME IS 3 E
TO THE POWER OF 3X.
NOW LET'S GO AHEAD AND CLEAN THIS UP A LITTLE BIT.
F PRIME OF X IS GOING TO BE EQUAL TO--
THIS WILL BE 15 E TO THE 3X x THE QUANTITY E
TO THE 3X - 1 TO THE 4TH POWER.
THIS ONE WAS SO MUCH FUN LET'S TRY ANOTHER.
HERE WE HAVE F OF X = THE SQUARE ROOT
OF 2 x U TO THE POWER X CUBED + 1.
SO THE FIRST THING WE NEED TO DO IS REWRITE THIS
AS A RATIONAL EXPONENT.
SO WE HAVE THE QUANTITY 2 E TO THE X TO THE 3RD POWER
+ 1 TO THE POWER OF 1/2.
THE NEXT THING WE NEED TO RECOGNIZE,
THIS IS A COMPOSITE FUNCTION WHERE U IS GOING TO BE
THE INNER FUNCTION.
SO U IS GOING TO BE EQUAL TO 2 E TO THE POWER OF X CUBED + 1.
AND THEN AS SOON AS WE IDENTIFY U WE KNOW WE NEED U PRIME.
TO DETERMINE U PRIME WE HAVE TO APPLY THE CHAIN RULE
WHEN WE DIFFERENTIATE E TO THE POWER OF X CUBED.
LOOKING AT OUR FORMULA HERE, WE NORMALLY LET THE EXPONENT
EQUAL U, BUT SINCE WE'VE ALREADY USED THE U HERE,
LET'S GO AHEAD AND LET X CUBED = V.
SO THE DERIVATIVE OF 2 x E TO THE V WOULD BE 2 x E
TO THE V x V PRIME.
WELL, IF V IS X CUBED, V PRIME WOULD BE 3X SQUARED
+ A DERIVATIVE OF 1, WHICH WOULD BE ZERO.
SO THIS WOULD BE 6X SQUARED E TO THE X CUBED.
NOW WE CAN DETERMINE F PRIME BECAUSE WE CAN THINK OF THIS
AS U TO THE 1/2.
SO THE DERIVATIVE OF U TO THE 1/2, WITH RESPECTS TO X,
WOULD BE 1/2 U TO THE -1/2 WHERE U IS 2 E TO THE X CUBED
+ 1 x U PRIME AND U PRIME IS 6X SQUARED E TO THE X CUBED.
AND ONE LAST STEP HERE TO CLEAN THIS UP.
1/2 x 6 THAT WOULD BE 3X SQUARED E TO THE X CUBED.
AND THIS IS A NEGATIVE EXPONENT, SO WE'LL MOVE IT DOWN
TO THE DENOMINATOR.
WE CAN WRITE THIS AS THE QUANTITY 2 E
TO THE X CUBED + 1 TO THE +1/2, OR, IF WE WANT,
WE CAN WRITE THIS AS A SQUARE ROOT.
WE'LL TAKE A LOOK AT SOME MORE EXAMPLES IN THE NEXT VIDEO.