- WELCOME TO IMPLICIT DIFFERENTIATION.
THE GOALS OF THIS VIDEO ARE TO DIFFERENTIATE IMPLICITLY,
AND ALSO TO FIND THE SLOPE OF A TANGENT LINE AT A GIVEN POINT.
EXPLICIT EQUATIONS OR FUNCTIONS ARE THOSE THAT ARE SOLVED FOR Y.
WE SAY THE VARIABLE Y IS EXPLICITLY WRITTEN
AS A FUNCTION OF X
WHEN IT IS SOLVED FOR Y AS WE SEE HERE.
IMPLICIT EQUATIONS OR FUNCTIONS ARE THOSE
THAT ARE NOT SOLVED FOR Y.
AS A RESULT WE MUST USE IMPLICIT DIFFERENTIATION
IF WE WANT TO FIND DYDX.
AN EXAMPLE OF AN IMPLICIT EQUATION WOULD LOOK LIKE THIS,
AND YOU COULD IMAGINE IF YOU TRIED TO SOLVE THIS FOR Y.
IT WOULD BE VERY CHALLENGING.
NOW, TO UNDERSTAND HOW TO FIND DYDX OF AN IMPLICIT EQUATION,
WE MUST REALIZE WE ARE TAKING THE DERIVATIVE
WITH RESPECTS TO X.
AS A RESULT, IF THE TERM ONLY INVOLVES X
WE WILL TAKE THE DERIVATIVE AS USUAL.
HOWEVER, IF THE TERM INVOLVES Y WE MUST APPLY THE CHAIN RULE,
BECAUSE WE ASSUME THAT Y IS A FUNCTION OF X,
OR Y IS DEFINED IMPLICITLY AS A DIFFERENTIABLE FUNCTION OF X.
SO LET'S TAKE A LOOK AT HOW WE ARE GOING TO DIFFERENTIATE
IMPLICITLY.
FIRST, WE WILL DIFFERENTIATE BOTH SIDES OF THE EQUATION
WITH RESPECTS TO X, OR POSSIBLY SOME OTHER VARIABLE.
THEN WE WILL APPLY THE RULES OF DIFFERENTIATION AS NECESSARY.
ANYTIME AN EXPRESSION INVOLVING Y IS DIFFERENTIATED,
DYDX WILL BE A FACTOR IN THE RESULT
BECAUSE OF THE CHAIN RULE.
NEXT WE'LL ISOLATE ALL TERMS WITH DYDX
AS A FACTOR ON ONE SIDE OF THE EQUATION.
THEN WE'LL FACTOR OUT THE DYDX,
AND THEN WE'LL DIVIDE ON BOTH SIDES OF THE EQUATION
TO ISOLATE OR SOLVE FOR DYDX.
LET'S GO AHEAD AND GIVE IT A TRY.
LET'S FIND DYDX,
THEN FIND THE SLOPE OF THE TANGENT LINE AT THE GIVEN POINT.
OKAY, SO STEP ONE IS TO DIFFERENTIATE
BOTH SIDES OF THE EQUATION WITH RESPECTS TO X.
SO IF THE TERM INVOLVES X WE'LL DIFFERENTIATE AS WE NORMALLY DO.
SO THE DERIVATIVE OF THIS TERM WOULD BE 4X -
NOW THIS TERM INVOLVES Y,
SO WE'LL DIFFERENTIATE THIS,
AND THEN HAVE A FACTOR OF DYDX.
SO THIS WOULD BE - 9Y SQUARED x DYDX.
AND AGAIN, WE'RE JUST THINKING OF Y AS SOME FUNCTION OF X,
AND WE'RE APPLYING THE CHAIN RULE.
THIS EQUALS 0, SO WE'LL ISOLATE THIS TERM ON ONE SIDE.
SO WE'LL SUBTRACT 4X ON BOTH SIDES,
AND THEN WE'LL DIVIDE BY -9Y SQUARED.
THESE SIMPLIFY, SO DYDX = 4X/9Y SQUARED.
SO THAT'S IMPLICIT DIFFERENTIATION.
NOW THE SECOND PART OF THIS QUESTION ASKS US
TO FIND THE SLOPE OF THE TANGENT LINE AT THE POINT -2,1.
SO NOW WE'RE GOING TO TAKE THE DERIVATIVE FUNCTION
AND EVALUATE IT AT THE POINT -2,1.
SO X IS -2 AND Y IS +1.
SO WE'LL HAVE 4 x X OR 4 x -2 / 9 x Y SQUARED OR 9 x 1 SQUARED,
WHICH WILL GIVE US -8/9.
SLOPE OF THE TANGENT LINE AT THE POINT -2,1 SHOULD BE -8/9.
LET'S GO AHEAD AND TAKE A LOOK AT THE GRAPH TO VERIFY THIS.
ONE THING WE'LL NOTICE IS WHEN WE GRAPH IMPLICIT EQUATIONS
WE OFTEN GET VERY INTERESTING GRAPHS.
HERE'S OUR POINT OF TANGENCY -2,1, AND WE CAN SEE A SLOPE.
WELL, THE SLOPE WAS -8/9, WHICH IS ALMOST -1,
AND WE CAN SEE THIS SLOPE DOES LOOK VERY CLOSE TO -1.
LET'S TRY ONE MORE.
FIRST STEP, WE NEED TO FIND DYDX,
SO WE NEED TO FIND THE DERIVATIVE WITH RESPECTS TO X
ON BOTH SIDES.
SO WHEN WE GO TO FIND THE DERIVATIVE OF Y CUBED
WE HAVE TO APPLY THE CHAIN RULE,
SO WE WOULD HAVE 3Y SQUARED x DYDX.
SOMETIMES YOU'LL SEE SOME TEXTBOOKS WRITING Y PRIME.
I DON'T LIKE DOING THAT,
BECAUSE IT LOOKS TOO MUCH LIKE Y TO THE 1st POWER.
NOW, PLUS THIS IS ACTUALLY A PRODUCT OF TWO FUNCTIONS.
WE HAVE TO APPLY THE PRODUCT RULE.
THE FIRST FUNCTION IS X SQUARED,
AND THE SECOND FUNCTION IS Y TO THE 5th.
SO WE'D HAVE X SQUARED x THE DERIVATIVE OF Y TO THE 5th,
AND AGAIN, WE HAVE TO APPLY THE CHAIN RULE
WHEN WE DIFFERENTIATE Y TO THE 5th.
SO WE'D HAVE 5Y TO THE 4th x DYDX
+ THE SECOND FUNCTION Y TO THE 5th x DERIVATIVE OF X SQUARED,
WHICH WOULD BE 2X - THE DERIVATIVE OF X TO THE 4th,
4X TO THE 3rd,
AND THE DERIVATIVE OF A CONSTANT OF COURSE IS 0.
NOW THIS NEXT STEP WE'RE GOING TO KEEP THE DYDX TERMS
ON THE LEFT SIDE,
AND WE'LL MOVE THE TERMS THAT DON'T HAVE DYDX IN THEM
TO THE RIGHT SIDE.
SO WHEN I REWRITE THE LEFT SIDE
I'M GOING TO FACTOR OUT THE DYDX AT THE SAME TIME.
SO I'D HAVE 3Y SQUARED + I'LL REWRITE THIS AS 5X SQUARED Y
TO THE 4th.
SO THESE ARE THE TERMS THAT HAD DYDX IN THEM,
WHICH MEANS I'LL BE MOVING THESE TERMS
TO THE RIGHT SIDE OF THE EQUATION AND HAVE A +4X CUBED,
AND THEN I'D HAVE A - 2XY TO THE 5th.
SINCE THESE ARE CONNECTED BY MULTIPLICATION,
THE LAST STEP WILL BE TO DIVIDE BY 3Y SQUARED + 5X SQUARED Y
TO THE 4th ON BOTH SIDES.
SO NOW WE SIMPLIFY, AND WE HAVE OUR DERIVATIVE.
OKAY, THE SECOND PART OF THIS QUESTION ASKS US
TO FIND THE SLOPE OF THE TANGENT LINE AT THE POINT 0,3.
SO WE NEED TO EVALUATE THIS DERIVATIVE
WHEN X IS 0 AND Y IS 3.
NOW WE MAY THINK THAT'S GOING TO BE A LOT OF WORK,
BUT SINCE X IS 0,
IF YOU TAKE A LOOK AT THIS NUMERATOR WE'D HAVE 4 x 0 CUBED,
THAT'S 0, - 2 x 0 x Y TO THE 5th.
THAT'S ALSO GOING TO BE 0, SO OUR NUMERATOR WOULD BE 0.
IN OUR DENOMINATOR WE'D HAVE 3 x 3 SQUARED + 5 x 0 SQUARED
x 3 TO THE 4th, OF COURSE IS 0.
SO THE VALUE OF THE DERIVATIVE AT THE POINT 0,3 = 0.
AGAIN, LET'S GO AHEAD AND LOOK AT A GRAPH TO VERIFY THIS.
HERE WE HAVE OUR FUNCTION.
HERE'S OUR POINT OF TANGENCY,
AND WE CAN EASILY SEE THAT AT THAT POINT
THE SLOPE OF THE TANGENT LINE IS 0.
AGAIN, THIS VERIFIES OUR WORK.
I HOPE THAT HELPED EXPLAIN HOW TO DIFFERENTIATE IMPLICITLY.
THANK YOU FOR WATCHING.