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X
- GIVEN F OF X EQUALS X RAISED TO THE POWER OF 2X
WE WANT TO FIND F PRIME OF X, THE DERIVATIVE OF F OF X,
AND F PRIME OF 1.
TO FIND OUR DERIVATIVE FUNCTION
WE'RE GOING TO PERFORM LOGARITHMIC DIFFERENTIATION.
SO FOR THE FIRST STEP WE'LL SUBSTITUTE Y FOR F OF X
SO OUR FUNCTION IS Y = X TO THE POWER OF 2X
AND NOW TO PERFORM LOG DIFFERENTIATION THE FIRST STEP
IS TO TAKE THE NATURAL LOG OF BOTH SIDES OF THE EQUATION.
NOW WE WANT TO EXPAND THE RIGHT SIDE OF THE EQUATION
AS MUCH AS POSSIBLE,
WHICH MEANS WE CAN USE THE POWER PROPERTY OF LOGARITHMS
TO MOVE THE EXPONENT TO THE FRONT
SO THAT WE HAVE A PRODUCT OF THE EXPONENT AND NATURAL LOG X.
SO NOW WE WOULD HAVE NATURAL LOG Y = 2X x NATURAL LOG X
AND NOW WE'LL DIFFERENTIATE BOTH SIDES OF THE EQUATION
WITH RESPECT TO X.
TO DIFFERENTIATE NATURAL LOG Y WITH RESPECT TO X
WE'LL PERFORM IMPLICIT DIFFERENTIATION WHICH MEANS
SINCE THIS FUNCTION IS IN TERMS OF Y NOT X
WE'LL HAVE AN EXTRA FACTOR OF DY/DX FROM THE CHAIN RULE.
SO THE DERIVATIVE OF NATURAL LOG Y WITH RESPECT TO X
WOULD BE 1/Y x DY/DX
AND ON THE RIGHT SIDE IN ORDER TO FIND THIS DERIVATIVE
WE'LL HAVE TO APPLY THE PRODUCT RULE.
IF WE NEED THE REVIEW, THE PRODUCT RULE IS GIVEN HERE
WHERE THE DERIVATIVE OF F x G WITH RESPECT TO X
IS EQUAL TO F x G PRIME + G x F PRIME.
SO IN THIS CASE, F WOULD BE EQUAL TO 2X
AND G WOULD BE EQUAL TO NATURAL LOG X.
SO THE DERIVATIVE IS EQUAL TO THE FIRST FUNCTION, 2X
x THE DERIVATIVE OF THE SECOND FUNCTION,
WHICH IS THE DERIVATIVE OF NATURAL LOG X
+ THE SECOND FUNCTION
WHICH IS NATURAL LOG X x THE DERIVATIVE OF THE FIRST FUNCTION
WHICH IS 2X.
NOW I'LL GO AHEAD AND FIND THESE TWO DERIVATIVES.
SO WE 1/Y x DY/DX = 2X x THE DERIVATIVE OF NATURAL LOG X
WITH RESPECT TO X WOULD JUST BE 1/X
+ NATURAL LOG X x THE DERIVATIVE OF 2X WHICH WOULD JUST BE 2.
LET'S GO AHEAD AND FIND THESE PRODUCTS.
WE HAVE 1/Y x DY/DX = NOTICE THESE X'S WOULD SIMPLIFY OUT.
WE JUST HAVE 2 + THIS WOULD BE 2 NATURAL LOG X.
NOW TO SOLVE THIS FOR DY/DX
WE'D HAVE TO MULTIPLY BOTH SIDES OF THE EQUATION BY Y.
SO WE'D HAVE DY/DX = Y x THE QUANTITY 2 + 2 NATURAL LOG X.
SO THIS IS OUR DERIVATIVE FUNCTION
BUT USUALLY IF WE CAN WE WANT TO WRITE THIS IN TERMS OF JUST X
NOT X, AND Y
AND SINCE Y OR OUR FUNCTION IS EQUAL TO X
RAISED TO THE POWER OF 2X
WE'LL SUBSTITUTE X TO THE POWER OF 2X FOR Y ON THE NEXT SLIDE.
SO AGAIN, WE SUBSTITUTED X TO THE POWER OF 2X FOR Y.
SO THIS IS OUR DERIVATIVE FUNCTION.
LET'S GO AHEAD AND DISTRIBUTE THIS.
SO WE HAVE DY/DX = 2 x X TO THE POWER OF 2X
+ 2X TO THE POWER OF 2X NATURAL LOG X.
REMEMBER WE WERE ALSO ASKED TO FIND F PRIME OF 1.
SO F PRIME 1 IS GOING TO BE EQUAL TO DY/DX
EVALUATED AT X = 1.
SO WE'D HAVE 2 x 1 TO THE POWER OF 2 x 1 THAT'S 2
+ 2 x 1 TO THE POWER OF 2 x NATURAL LOG 1
BUT NATURAL LOG 1 IS EQUAL TO ZERO.
SO THIS SIMPLIFIES TO ZERO.
SO F PRIME OF 1 IS EQUAL TO +2
WHICH IF WE GRAPH THE FUNCTION
WOULD BE THE SLOPE OF THE TANGENT LINE AT X = 1.
SO BEFORE WE LOOK AT THE GRAPH,
AGAIN HERE WAS OUR DERIVATIVE FUNCTION F PRIME OF X OR DY/DX
AND HERE'S F PRIME OF 1.
SO IF WE TAKE A LOOK AT THE GRAPH
OF THE ORIGINAL FUNCTION F OF X,
HERE'S THE POINT ON THE FUNCTION WHEN X = +1
AND IF WE SKETCH A TANGENT LINE AT THAT POINT
IT WOULD LOOK SOMETHING LIKE THIS
AND SINCE F PRIME OF 1 WAS EQUAL TO +2
WE KNOW THE SLOPE OF THIS TANGENT LINE AT THAT POINT
IS +2.
OKAY. I HOPE YOU FOUND THIS EXPLANATION HELPFUL.