- WELCOME TO A SECOND VIDEO ON THE CHAIN RULE
FOR FUNCTIONS OF TWO VARIABLES.
IN THIS VIDEO WE'LL DETERMINE THE PARTIAL DERIVATIVES
OF A FUNCTION OF TWO VARIABLES WITH TWO INDEPENDENT VARIABLES.
IN THE PREVIOUS VIDEO WE DISCUSSED HOW TO DETERMINE
DZDT WITH A FUNCTION IN TWO VARIABLES
WITH ONE INDEPENDENT VARIABLE AS WE SEE HERE.
BUT NOW WE'RE GOING TO TAKE A LOOK AT DETERMINING
THE PARTIAL DERIVATIVE OF Z WHEN Z IS DEFINED
IN TERMS OF X AND Y, BUT X IS IN TERMS
OF TWO VARIABLES AND SO IS Y.
SO TO DIAGRAM THIS, WE START WITH FUNCTION Z
THAT'S DEFINED IN TERMS OF X AND Y.
BUT THEN X AND Y ARE DEFINED IN TERMS OF S AND T.
WE CAN USE THIS TREE DIAGRAM TO HELP US REMEMBER THE FORMULA
FOR THE CHAIN RULE.
MEANING IF WE WANT THE PARTIAL OF Z WITH RESPECTS TO S
WE NEED TO FOLLOW ALL OF THE PATHS
DOWN TO THE BOTTOM THAT END IN S.
SO THE PARTIAL Z WITH RESPECTS TO X WOULD BE THIS BRANCH HERE.
NOTICE HOW WE GO FROM Z TO X.
AND NOW WE HAVE TO GO FROM X TO S WHICH WOULD BE THE PARTIAL
OF X WITH RESPECTS TO S AS WE SEE HERE IN THE FORMULA.
AND, NOW, THERE'S ALSO A PATH TO S TO THE RIGHT.
THE FIRST PATH LEADS FROM Z TO Y THAT REPRESENTS THE PARTIAL OF Z
WITH RESPECTS TO Y AS WE SEE HERE.
AND THEN FROM Y TO S THAT WOULD BE THE PARTIAL OF Y
WITH RESPECTS TO S AS WE SEE HERE.
AND THEN TO DETERMINE THE PARTIAL DERIVATIVE OF Z
WITH RESPECTS TO T WE WOULD JUST FOLLOW--MEANING WE'D GO
FROM HERE TO HERE, THIS WOULD BE THIS HALF OF THE CHAIN RULE.
MEANING ON THE RIGHT WE WOULD GO FROM Z TO Y AND THEN Y TO T,
WHICH REPRESENTS THE SECOND HALF OF THE CHAIN RULE.
LET'S GO AND TAKE A LOOK AT SOME EXAMPLES.
HERE WE WANT TO FIND THE PARTIAL DERIVATIVE
WITH RESPECTS TO S AND T.
LET'S START BY FINDING THE PARTIAL OF Z WITH RESPECTS TO S.
SO WE'RE GOING TO FIRST FIND THE PARTIAL Z
WITH RESPECTS TO X, TREATING Y AS A CONSTANT.
SO THE DERIVATIVE OF SINE 2X IS GOING TO BE COSINE 2X x 2.
SO WE'LL HAVE 2 COSINE 2X AND THEN x THE CONSTANT COSINE
Y x THE PARTIAL DERIVATIVE OF X WITH RESPECTS TO S.
WELL, HERE'S X, TREATING T AS A CONSTANT,
WE WOULD JUST HAVE 2.
+ THE PARTIAL OF Z WITH RESPECTS TO Y,
SO SINE 2X IS NOW OUR CONSTANT.
THE DERIVATIVE OF COSINE Y IS GOING TO BE -SINE Y.
SO NOW LET'S MULTIPLY THESE TOGETHER
AND THEN REPLACE X WITH 2S + T AND Y WITH S - 2T.
SO WE HAVE 2 x 2 THAT'S 4.
AND THEN FOR COSINE 2X WE'LL HAVE 2 x 2S + T
THAT'LL BE 4S + 2T.
AND THEN WE HAVE COSINE Y, WELL, COSINE Y
WILL BE COSINE OF S - 2T.
OVER HERE IT'S GOING TO END UP BEING MINUS.
WE'D HAVE SINE 2X SINE Y, SO WE'LL HAVE SINE OF 2 x 2S + T
THAT'S 4S + 2T x SINE Y AND Y IS S - 2T.
NOW WE'LL GO TO THE NEXT SLIDE AND FIND THE PARTIAL DERIVATIVE
OF Z WITH RESPECTS TO T.
SO FOR THE PARTIAL OF Z WITH RESPECTS TO T
WE'RE GOING TO HAVE THE PARTIAL OF Z WITH RESPECTS TO X AGAIN.
THE DERIVATIVE OF SINE 2X THAT'S GOING TO BE COSINE 2X x 2
OR 2 COSINE 2X.
AND COSINE Y WOULD BE OUR CONSTANT.
AGAIN, WE HAD THIS ON THE LAST SCREEN.
x THE PARTIAL OF X WITH RESPECTS TO T IS JUST GOING TO BE 1.
+ THE PARTIAL DERIVATIVE OF Z WITH RESPECTS TO Y,
SO SINE 2X IS A CONSTANT, AND THE DERIVATIVE
OF COSINE Y IS GOING TO BE -SINE Y.
SO WE HAVE SINE 2X x -SINE Y.
AND THEN WE HAVE THE PARTIAL OF Y WITH RESPECTS TO T.
WELL, HERE'S Y SO S IS A CONSTANT.
DERIVATIVE OF -2T WOULD BE -2.
LET'S GO AHEAD AND REWRITE THIS AND ALSO REPLACE X
WITH 2S + T AND Y WITH S - 2T.
SO HERE WE'LL HAVE 2 x COSINE 2X.
WELL, IF X IS 2S + T, 2X WILL BE 4S + 2T.
AND THEN COSINE Y IS GOING TO BE THE COSINE OF S - 2T.
NOW, HERE WE HAVE A PRODUCT OF TWO NEGATIVES,
SO WE'RE GOING TO HAVE + SINE 2X IS GOING
TO BE THE SINE OF 4S + 2T.
AND THEN SINE Y IS GOING TO BE SINE OF S - 2T.
AND THAT'S GOING TO DO IT FOR THIS PROBLEM.
LET'S GO AND TAKE A LOOK AT ONE MORE EXAMPLE.
HERE'S OUR FUNCTION Z IN TERMS OF X AND Y.
HERE'S X AND HERE'S Y.
SO LET'S START WITH THE PARTIAL OF Z WITH RESPECTS TO S.
SO WE'LL START BY DETERMINING THE PARTIAL OF Z
WITH RESPECTS TO X.
SO WE'RE TREATING Y AS A CONSTANT.
WE'D HAVE E TO THE XY x THE DERIVATIVE OF XY
WOULD BE Y x THE PARTIAL OF X WITH RESPECTS TO S.
WELL, HERE'S X.
SO DERIVATIVE OF S SQUARE WOULD BE 2S + THIS WOULD BE 2T.
+ THE PARTIAL OF Z WITH RESPECTS TO Y.
NOW WE'RE GOING TO HAVE E TO THE XY x X x THE PARTIAL
OF Y WITH RESPECTS TO S.
SO THAT'S A CONSTANT.
THIS WOULD BE 2T.
FOR SAKE OF TIME I'M GOING TO GO AHEAD AND LEAVE IT LIKE THIS.
LET'S CLEAN THIS UP A LITTLE BIT.
I'M NOT GOING TO PERFORM THE SUBSTITUTIONS THIS TIME.
WE'LL HAVE Y E TO THE XY x 2S + 2T
+ THIS WILL BE X E TO THE XY x 2T.
AGAIN, WE COULD PERFORM THE SUBSTITUTIONS HERE,
BUT IT WOULD PRETTY MESSY AND A LOT OF ALGEBRA.
THE WHOLE POINT HERE IS TO GET PRACTICE WITH THE CHAIN RULE.
SO LET'S GO AHEAD AND DETERMINE THE PARTIAL DERIVATIVE OF Z
WITH RESPECTS TO T.
IT'S NICE TO HAVE IT ON THE SAME SCREEN BECAUSE WE ARE GOING
TO SEE SOME REPETITIONS HERE IN THESE PARTIAL DERIVATIVES.
MEANING THE FIRST PARTIAL IS THE PARTIAL DERIVATIVE OF Z
WITH RESPECTS TO X, WHICH WE FOUND OVER HERE,
WHICH WE FOUND TO BE E TO THE XY x Y.
x THE PARTIAL DERIVATIVE OF X WITH RESPECTS TO T.
WELL, HERE'S X, SO THAT'S A CONSTANT.
SO WE'D HAVE 2S + THE PARTIAL OF Z WITH RESPECTS TO Y.
AGAIN, WE FOUND THAT OVER HERE.
IT WAS E TO THE XY x X.
AND THEN THE PARTIAL OF Y WITH RESPECTS TO T.
WELL, HERE'S Y, S IS NOW A CONSTANT,
SO WE'D HAVE 2T + 2S.
AGAIN, WE'RE GOING TO LEAVE IT IN TERMS OF X IS Y,
AS S AS IN T's. LET'S CLEAN IT UP A LITTLE BIT.
WE COULD REWRITE THIS AS Y E TO THE XY 2S + X E
TO THE XY x 2T + 2S.
I'M GOING TO GO AHEAD AND STOP HERE FOR THIS VIDEO.
I HOPE YOU FOUND THESE EXAMPLES HELPFUL.