- WELCOME TO A VIDEO ON FIRST PARTIAL DERIVATIVES.
LET'S START WITH A QUICK REVIEW.
THE FIRST DERIVATIVE OF A FUNCTION OF ONE VARIABLE
TELLS US THE SLOPE OF THE TANGENT LINE AT A GIVEN POINT.
THIS IS THE RATE OF CHANGE OF Y WITH RESPECTS TO X.
THE FIRST PARTIAL DERIVATIVE OF A FUNCTION OF TWO VARIABLES
TELLS US THE SLOPE OF THE TANGENT LINE AT A GIVEN POINT
IN EITHER THE X DIRECTION OR Y DIRECTION.
THEREFORE, THE PARTIAL DERIVATIVE MEASURES
THE RATE OF CHANGE OF Z WITH RESPECTS TO X
OR THE RATE OF CHANGE OF Z WITH RESPECTS TO Y.
HERE WE SEE THE LIMIT DEFINITIONS
OF THE PARTIAL DERIVATIVE WITH RESPECTS TO X HERE
AND WITH RESPECTS TO Y HERE,
BUT WE'RE NOT GOING TO DETERMINE
PARTIAL DERIVATIVES IN THIS VIDEO IN THIS WAY.
IN THIS VIDEO,
TO DETERMINE THE PARTIAL DERIVATIVE OF F
WITH RESPECTS TO X AS WE SEE HERE,
WE'LL CONSIDER Y A CONSTANT AND DIFFERENTIATE WITH RESPECTS TO X
AND TO FIND THE PARTIAL DERIVATIVE
OF F WITH RESPECTS TO Y,
WE'LL CONSIDER X A CONSTANT AND DIFFERENTIATE WITH RESPECTS TO Y
BEFORE WE DO THAT,
LET'S GO AHEAD AND TAKE A LOOK AT SOME DIFFERENT NOTATION
FOR PARTIAL DERIVATIVES.
ALL OF THESE NOTATIONS HERE REPRESENT
PARTIAL DERIVATIVES OF F WITH RESPECTS TO X
OR WITH RESPECTS TO Y.
AND ONE THEY LEFT OFF HERE IS THE PARTIAL OF F
INSTEAD OF THE PARTIAL OF Z WITH RESPECTS TO X
AND THE PARTIAL OF F WITH RESPECTS TO Y.
AND IF YOU WANT TO EVALUATE THESE AT A SPECIFIC POINT,
WE CAN USE THE NOTATION AS WE SEE HERE DOWN BELOW.
LET'S GO AHEAD AND TAKE A LOOK AT SOME EXAMPLES
AND THEN LOOK AT THE RESULTS GEOMETRICALLY.
HERE WE WANT TO DETERMINE THE PARTIAL DERIVATIVE
WITH RESPECTS TO X AND Y AT THE POINT (1, 2).
SO TO FIND THE PARTIAL DERIVATIVE OF F
WITH RESPECTS TO X,
WE'RE GOING TO DIFFERENTIATE WITH RESPECTS TO X,
TREATING Y AS A CONSTANT.
SO THE DERIVATIVE OF X SQUARED WOULD BE 2X.
THE DERIVATIVE OF Y SQUARED WOULD BE ZERO,
BECAUSE WE'RE TREATING IT AS A CONSTANT.
DERIVATIVE OF FOUR IS ZERO.
NOW, WE'LL GO AHEAD AND EVALUATE
THIS PARTIAL DERIVATIVE AT THE POINT (1, 2).
WELL, THIS ONLY REQUIRES THE VALUE OF X. SO WE'LL HAVE 2 x 1.
THAT'LL GIVE US A VALUE OF 2.
AT THE POINT (1, 2),
THE SLOPE OF THE TANGENT LINE IN THE X DIRECTION = 2.
NOW, THE PARTIAL OF F WITH RESPECTS TO Y--
SO NOW, WE'LL DIFFERENTIATE WITH RESPECTS TO Y
AND TREAT X AS A CONSTANT. SO THAT DERIVATIVE WOULD BE ZERO.
THIS'LL BE -2Y AND THIS WOULD BE ZERO.
SO IF WE EVALUATE THIS AT THE POINT (1, 2), Y = 2.
THIS WOULD BE -4.
THE SLOPE OF THE TANGENT LINE
AT THIS POINT IN THE Y DIRECTION SHOULD BE -4.
LET'S TAKE A LOOK AT THAT GRAPHICALLY.
WHAT WE'RE SEEING HERE IS THE SURFACE IN BLUE
AND THE TANGENT LINE IN BLACK
AND THE POINT OF TANGENT C IN RED,
AND IF I ANGLE THIS CORRECTLY,
NOTICE AS WE MOVE OUT ALONG THE POSITIVE X AXIS,
THE LINE IS MOVING UP,
AND THAT SLOPE IN THE X DIRECTION WOULD BE +2.
NOW, LET'S TAKE A LOOK
AT THE SLOPE OF THE TANGENT LINE IN THE Y DIRECTION.
NOTICE NOW, AS WE MOVE RIGHT ALONG THE POSITIVE Y AXIS,
THE LINE IS MOVING DOWNWARD,
AND THE SLOPE OF THAT LINE IN THE Y DIRECTION IS -4.
SO THIS IS A NICE ILLUSTRATION
OF WHAT WE'RE DETERMINING WHEN WE FIND PARTIAL DERIVATIVES.
LET'S GO AHEAD AND TAKE A LOOK AT ANOTHER EXAMPLE.
NOW, NOTICE IN THIS FUNCTION THE FIRST THING WE CAN DO
IS SIMPLIFY THIS AND THEN APPLY A PROPERTY OF LOGARITHMS
TO HELP FINDING THESE PARTIAL DERIVATIVES A LITTLE BIT EASIER.
WHAT I MEAN BY THAT IS THIS IS THE SAME AS NATURAL LOG
OF XY TO THE POWER OF 1/2,
AND THEN WHAT WE CAN DO IS TAKE ADVANTAGE
OF THE POWER PROPERTY OF LOGS
AND MOVE THIS TO THE POSITION OF THE COEFFICIENT.
SO WE'RE GOING TO USE THE FUNCTION
F OF XY = 1/2 NATURAL LOG XY.
ONE MORE THING WE SHOULD REMEMBER,
THE DERIVATIVE OF NATURAL LOG U = 1/U x U PRIME. OKAY.
LET'S GO AHEAD
AND FIND THE PARTIAL DERIVATIVE WITH RESPECTS TO X.
SO WE'RE GOING TO HAVE 1/2 x 1/U x U PRIME.
SO 1/XY--NOW, U PRIME--
WE'LL FIND THE DERIVATIVE WITH RESPECTS TO X,
TREATING Y AS A CONSTANT. THE DERIVATIVE OF X IS ONE.
SO WE'LL HAVE 1 x Y OR JUST Y. THIS Y AND THIS Y SIMPLIFY OUT.
SO THE PARTIAL WITH RESPECTS TO X IS JUST 1/2X.
SO IF WE EVALUATE THIS AT THE POINT (2, 2),
OF COURSE, X = 2 SO WE'LL HAVE 1/4.
NOW, LET'S DETERMINE THE PARTIAL WITH RESPECTS TO Y.
IT'S GOING TO BE A SIMILAR PROCESS.
WE'LL HAVE 1/2 x 1/U, 1/XY x U PRIME,
BUT NOW WE'RE TREATING X AS A CONSTANT
AND DIFFERENTIATING WITH RESPECTS TO Y.
SO WE'LL HAVE X x 1 OR JUST X. SO NOW, THE Xs SIMPLIFY OUT,
AND WE'RE LEFT WITH 1/2Y.
SO AT THE POINT (2, 2), WE'RE GOING TO HAVE 1/2 x 2 OR 1/4.
SO THE SLOPE IN THE X DIRECTION AND THE Y DIRECTION
ARE THE SAME AT THE POINT (2, 2).
LET'S LOOK AT THIS ONE GRAPHICALLY AS WELL.
WE SAID THE SLOPE IN THE X AND Y DIRECTION WOULD BE +1/4.
IF WE LOOK AT THIS ALONG THE POSITIVE X AXIS,
YOU CAN SEE IT'S MOVING UP
FROM LEFT TO RIGHT GRADUALLY AT THAT POINT.
THIS MODEL IS THE PARTIAL DERIVATIVE WITH RESPECTS TO X,
AND HERE'S THE MODEL OF THE PARTIAL DERIVATIVE
WITH RESPECTS TO Y. HERE'S THE POSITIVE Y AXIS.
IF WE ANGLE THIS CORRECTLY,
AS WE MOVE ALONG THE POSITIVE Y AXIS,
THAT TANGENT LINE IS MOVING UPWARD WITH A SLOPE OF 1/4.
LET'S GO AHEAD AND TAKE A LOOK AT A COUPLE MORE.
HERE WE WANT TO FIND THE PARTIAL DERIVATIVE
WITH RESPECTS TO X AND Y
AND THEN EVALUATE THEM AT THE POINT (2, -1).
THIS IS GOING TO REQUIRE THE QUOTIENT RULE.
LET'S START BY DETERMINING
THE PARTIAL DERIVATIVE WITH RESPECTS TO X.
SO OUR DENOMINATOR IS GOING TO BE X + Y SQUARED,
AND THEN WE'RE GOING TO HAVE
THE DENOMINATOR x THE DERIVATIVE OF THE NUMERATOR.
SO WE'LL HAVE X + Y.
REMEMBER, WE'RE TREATING Y AS A CONSTANT.
SO THE DERIVATIVE OF XY WOULD JUST BE 1 x Y, OR Y,
- THE NUMERATOR x THE DERIVATIVE OF THE DENOMINATOR
WITH RESPECTS TO X, TREATING Y AS A CONSTANT.
SO THIS WOULD BE 1, THE DERIVATIVE OF Y WOULD BE 0.
LET'S GO AHEAD AND SIMPLIFY THIS.
IF WE DISTRIBUTE HERE,
WE'RE GOING TO HAVE XY + Y SQUARED AND THEN - XY.
SO THE XY TERMS ARE OPPOSITES.
WE'RE GOING TO HAVE Y SQUARED /X + Y TO THE 2nd.
LET'S GO AHEAD AND EVALUATE THIS AT THE POINT (2, -1).
SO OUR NUMERATOR'S GOING TO BE 1,
DENOMINATOR WILL BE X + Y. THAT'S GOING TO BE 1 SQUARED.
THAT'S 1-- I DIDN'T LEAVE ENOUGH ROOM--
SO I'LL TRY TO SQUEEZE IT IN DOWN HERE.
THE PARTIAL DERIVATIVE WITH RESPECTS TO Y IS GOING TO BE
THE DERIVATIVE OF THIS WITH RESPECTS TO Y,
TREATING X AS A CONSTANT.
SO WE'LL HAVE X + Y SQUARED, APPLYING THE QUOTIENT RULE.
THEN WE'LL HAVE X + Y x THE DERIVATIVE OF XY,
TREATING X AS A CONSTANT,
SO IT WILL BE X x THE DERIVATIVE OF Y WOULD BE 1.
SO WE'LL HAVE X - XY NUMERATOR
x THE DERIVATIVE OF THE DENOMINATOR,
TREATING X AS A CONSTANT. THAT'D BE 0 AND THAT WOULD BE 1.
WHEN WE DISTRIBUTE HERE,
WE'RE GOING TO HAVE X SQUARED + XY - XY.
SO THOSE TWO TERMS ARE OPPOSITES.
WE'RE LEFT WITH X SQUARED /THE QUANTITY X + Y SQUARED,
AND NOW, WE NEED TO EVALUATE THIS AT THE POINT (2, -1).
SO OUR DENOMINATOR IS GOING TO BE X + Y SQUARED.
THAT'S GOING TO BE 1 SQUARED OR 1.
OUR NUMERATOR IS GOING TO BE X SQUARED. 2 SQUARED = 4.
SO THIS TELLS US AT THE POINT (2, -1),
THE SLOPE OF THE TANGENT LINE IN THE X DIRECTION SHOULD BE +1,
AND THE SLOPE IN THE Y DIRECTION SHOULD BE +4.
LET'S TRY ONE MORE.
HERE WE WANT THE PARTIAL DERIVATIVE
WITH RESPECTS TO X AND Y.
WE HAVE F OF XY = SINE X SQUARED + Y SQUARED.
SO X SQUARED + Y SQUARED = U.
SO THE PARTIAL DERIVATIVE WITH RESPECTS TO X--
WELL, THE DERIVATIVE OF SINE U IS COSINE U.
SO WE'LL HAVE COSINE OF X SQUARED + Y SQUARED x U PRIME.
SO THE DERIVATIVE OF X SQUARED WOULD BE 2X.
THE DERIVATIVE OF Y SQUARED WOULD BE 0,
BECAUSE WE'RE TREATING IT LIKE A CONSTANT.
SO OUR PARTIAL DERIVATIVE WITH RESPECTS TO X
IS JUST 2X COSINE X SQUARED + Y SQUARED.
NOW, LET'S FIND THE PARTIAL WITH RESPECTS TO Y.
IT'S GOING TO BE COSINE U x U PRIME AGAIN,
BUT NOW, U PRIME--
THE DERIVATIVE WITH RESPECTS TO Y--THIS WOULD BE 0,
AND THIS WOULD BE 2Y.
NOW, THAT'S ALL WE HAVE TIME FOR IN THIS VIDEO.
I HOPE YOU FOUND THESE EXAMPLES HELPFUL.