- WELCOME TO THE PRODUCT RULE.
- WELCOME TO THE PRODUCT RULE.

THE GOAL OF THIS VIDEO WILL BE TO FIND DERIVATIVES
THE GOAL OF THIS VIDEO WILL BE TO FIND DERIVATIVES

USING THE PRODUCT RULE AND ALSO TO FIND THE EQUATION
USING THE PRODUCT RULE AND ALSO TO FIND THE EQUATION

OF A TANGENT LINE.
OF A TANGENT LINE.

LET'S GO AHEAD AND GET STARTED.
LET'S GO AHEAD AND GET STARTED.

WE'VE ALREADY FOUND THAT THE DERIVATIVE A SUM IS THE SUM
WE'VE ALREADY FOUND THAT THE DERIVATIVE A SUM IS THE SUM

OF THE DERIVATIVES AS STATED HERE.
OF THE DERIVATIVES AS STATED HERE.

HOWEVER, THE DERIVATIVES OF A PRODUCT
HOWEVER, THE DERIVATIVES OF A PRODUCT

IS NOT THE PRODUCT OF THE DERIVATIVES.
IS NOT THE PRODUCT OF THE DERIVATIVES.

LET'S TAKE A LOOK AT WHY IT'S NOT THE PRODUCT
LET'S TAKE A LOOK AT WHY IT'S NOT THE PRODUCT

OF THE DERIVATIVES.
OF THE DERIVATIVES.

WE KNOW THE DERIVATIVE OF X TO THE 5th
WE KNOW THE DERIVATIVE OF X TO THE 5th

BY USING THE POWER RULE IS EQUAL TO 5X TO THE 4th.
BY USING THE POWER RULE IS EQUAL TO 5X TO THE 4th.

SO IF WE TRY TO FIND THE PRODUCT OF DERIVATIVES,
SO IF WE TRY TO FIND THE PRODUCT OF DERIVATIVES,

IT'S EASY TO SEE THIS DOES NOT WORK.
IT'S EASY TO SEE THIS DOES NOT WORK.

FOR EXAMPLE, INSTEAD OF FINDING THE DERIVATIVE OF X TO THE 5th,
FOR EXAMPLE, INSTEAD OF FINDING THE DERIVATIVE OF X TO THE 5th,

LET'S SAY I HAVE A DERIVATIVE OF X SQUARED x X TO THE 3rd.
LET'S SAY I HAVE A DERIVATIVE OF X SQUARED x X TO THE 3rd.

SO HERE WE HAVE A PRODUCT.
SO HERE WE HAVE A PRODUCT.

IF WE TAKE THE DERIVATIVE OF EACH OF THESE
IF WE TAKE THE DERIVATIVE OF EACH OF THESE

AND THEN MULTIPLY THEM,
AND THEN MULTIPLY THEM,

HERE WE'D HAVE 2X AND THEN WE'D HAVE 3X SQUARED.
HERE WE'D HAVE 2X AND THEN WE'D HAVE 3X SQUARED.

OBVIOUSLY THAT'S EQUAL TO 6X CUBED
OBVIOUSLY THAT'S EQUAL TO 6X CUBED

WHICH OF COURSE IS NOT EQUAL TO 5X TO THE 4th,
WHICH OF COURSE IS NOT EQUAL TO 5X TO THE 4th,

SO THIS JUST DOESN'T WORK.
SO THIS JUST DOESN'T WORK.

WE CANNOT FIND THE DERIVATIVES OF A PRODUCT
WE CANNOT FIND THE DERIVATIVES OF A PRODUCT

BY FINDING THE PRODUCT OF THE DERIVATIVES.
BY FINDING THE PRODUCT OF THE DERIVATIVES.

WE'RE GOING TO HAVE TO FIND A DIFFERENT METHOD,
WE'RE GOING TO HAVE TO FIND A DIFFERENT METHOD,

SO THIS WILL NOT WORK.
SO THIS WILL NOT WORK.

AND THE WAY WE FIND THE DERIVATIVE OF A PRODUCT
AND THE WAY WE FIND THE DERIVATIVE OF A PRODUCT

IS FROM THE PRODUCT RULE, AND HERE'S WHAT IT SAYS.
IS FROM THE PRODUCT RULE, AND HERE'S WHAT IT SAYS.

THE DERIVATIVE OF A PRODUCT IS THE FIRST FACTOR, F OF X,
THE DERIVATIVE OF A PRODUCT IS THE FIRST FACTOR, F OF X,

x THE DERIVATIVE OF THE SECOND FACTOR,
x THE DERIVATIVE OF THE SECOND FACTOR,

SO G PRIME OF X + THE SECOND FACTOR, G OF X,
SO G PRIME OF X + THE SECOND FACTOR, G OF X,

x THE DERIVATIVE OF THE FIRST FACTOR
x THE DERIVATIVE OF THE FIRST FACTOR

WHICH WOULD BE F PRIME OF X.
WHICH WOULD BE F PRIME OF X.

SO WE HAVE THE FIRST x THE DERIVATIVE OF THE SECOND
SO WE HAVE THE FIRST x THE DERIVATIVE OF THE SECOND

+ THE SECOND x THE DERIVATIVE OF THE FIRST.
+ THE SECOND x THE DERIVATIVE OF THE FIRST.

LET'S GO AHEAD AND APPLY THIS TO THE PREVIOUS EXAMPLE.
LET'S GO AHEAD AND APPLY THIS TO THE PREVIOUS EXAMPLE.

AGAIN, WE KNOW THAT THE DERIVATIVE OF X TO THE 5th
AGAIN, WE KNOW THAT THE DERIVATIVE OF X TO THE 5th

IS 5X TO THE 4th.
IS 5X TO THE 4th.

SO IF WE BREAK IT UP INTO A PRODUCT
SO IF WE BREAK IT UP INTO A PRODUCT

AND THEN WE APPLY THE PRODUCT RULE,
AND THEN WE APPLY THE PRODUCT RULE,

LET'S SEE IF WE GET THE SAME THING.
LET'S SEE IF WE GET THE SAME THING.

SO HERE WE HAVE THE FIRST FUNCTION
SO HERE WE HAVE THE FIRST FUNCTION

x THE DERIVATIVE OF THE SECOND,
x THE DERIVATIVE OF THE SECOND,

WE'RE GOING TO COME BACK AND FIND THAT,
WE'RE GOING TO COME BACK AND FIND THAT,

+ THE SECOND x THE DERIVATIVE OF THE FIRST.
+ THE SECOND x THE DERIVATIVE OF THE FIRST.

I HAVEN'T DONE ANY CALCULUS YET,
I HAVEN'T DONE ANY CALCULUS YET,

I'VE JUST WRITTEN OUT WHAT WE'RE GOING TO DO.
I'VE JUST WRITTEN OUT WHAT WE'RE GOING TO DO.

SO NOW, WE'RE GOING TO FIND THE DERIVATIVE OF X TO THE 3rd
SO NOW, WE'RE GOING TO FIND THE DERIVATIVE OF X TO THE 3rd

WHICH OF COURSE IS 3X SQUARED
WHICH OF COURSE IS 3X SQUARED

+ X TO THE 3rd x THE DERIVATIVE OF X SQUARED, THAT WOULD BE 2X.
+ X TO THE 3rd x THE DERIVATIVE OF X SQUARED, THAT WOULD BE 2X.

SO SIMPLIFYING HERE, WE WOULD HAVE 3X TO THE 4th
SO SIMPLIFYING HERE, WE WOULD HAVE 3X TO THE 4th

+ 2X TO THE 4th, AND THIS DOES VERIFY
+ 2X TO THE 4th, AND THIS DOES VERIFY

WE WOULD HAVE 5X TO THE 4th
WE WOULD HAVE 5X TO THE 4th

WHICH WE KNOW TO BE THE DERIVATIVE.
WHICH WE KNOW TO BE THE DERIVATIVE.

NOW OF COURSE, THIS ISN'T A PROOF.
NOW OF COURSE, THIS ISN'T A PROOF.

THE PROOF IS IN THE TEXT.
THE PROOF IS IN THE TEXT.

BUT THIS DOES VERIFY THAT THIS FORMULA DOES WORK
BUT THIS DOES VERIFY THAT THIS FORMULA DOES WORK

AT LEAST IN THIS CASE, AND IN FACT IT WORKS FOR ANY PRODUCT.
AT LEAST IN THIS CASE, AND IN FACT IT WORKS FOR ANY PRODUCT.

LET'S APPLY IT TO THIS FUNCTION.
LET'S APPLY IT TO THIS FUNCTION.

NOW OVER HERE, I'VE WRITTEN THE PRODUCT RULE
NOW OVER HERE, I'VE WRITTEN THE PRODUCT RULE

SO WE CAN REFERENCE IT, AND I'VE SHORTENED IT A LITTLE BIT.
SO WE CAN REFERENCE IT, AND I'VE SHORTENED IT A LITTLE BIT.

I'VE JUST CALLED THE FIRST FUNCTION "F"
I'VE JUST CALLED THE FIRST FUNCTION "F"

AND THE SECOND FUNCTION "G," BUT IT'S THE SAME FORMULA.
AND THE SECOND FUNCTION "G," BUT IT'S THE SAME FORMULA.

SO IN THIS PROBLEM HERE,
SO IN THIS PROBLEM HERE,

THIS IS OUR FUNCTION F AND THIS IS OUR FUNCTION G.
THIS IS OUR FUNCTION F AND THIS IS OUR FUNCTION G.

LET'S WRITE OUT WHAT WE HAVE TO FIND AND THEN WE'LL FIND IT.
LET'S WRITE OUT WHAT WE HAVE TO FIND AND THEN WE'LL FIND IT.

SO THE DERIVATIVE IS EQUAL TO THE FIRST FUNCTION
SO THE DERIVATIVE IS EQUAL TO THE FIRST FUNCTION

x THE DERIVATIVE OF THE SECOND + THE SECOND FUNCTION
x THE DERIVATIVE OF THE SECOND + THE SECOND FUNCTION

x THE DERIVATIVE OF THE FIRST FUNCTION.
x THE DERIVATIVE OF THE FIRST FUNCTION.

I HAVEN'T FOUND THESE DERIVATIVES YET.
I HAVEN'T FOUND THESE DERIVATIVES YET.

I JUST WROTE OUT THE PRODUCT RULE,
I JUST WROTE OUT THE PRODUCT RULE,

AND NOW WE WILL CALCULATE THE DERIVATIVES.
AND NOW WE WILL CALCULATE THE DERIVATIVES.

THE DERIVATIVE OF 2X SQUARED
THE DERIVATIVE OF 2X SQUARED

+ 3X + 5 WOULD BE 4X + 3 + THE SECOND FUNCTION
+ 3X + 5 WOULD BE 4X + 3 + THE SECOND FUNCTION

x THE DERIVATIVE OF 4X - 3 THAT WOULD BE 4.
x THE DERIVATIVE OF 4X - 3 THAT WOULD BE 4.

SO THIS IS OUR DERIVATIVE FUNCTION.
SO THIS IS OUR DERIVATIVE FUNCTION.

NOW, WE JUST HAVE TO CLEAN IT UP,
NOW, WE JUST HAVE TO CLEAN IT UP,

MEANING WE HAVE TO MULTIPLY ALL THIS OUT
MEANING WE HAVE TO MULTIPLY ALL THIS OUT

AND THEN COMBINE LIKE TERMS.
AND THEN COMBINE LIKE TERMS.

SO TO SAVE TIME, I MULTIPLIED ALL THIS OUT
SO TO SAVE TIME, I MULTIPLIED ALL THIS OUT

AND I'VE IDENTIFIED THE LIKE TERMS.
AND I'VE IDENTIFIED THE LIKE TERMS.

NOW, I'M GOING TO COMBINE THE LIKE TERMS
NOW, I'M GOING TO COMBINE THE LIKE TERMS

TO FIND THE FINAL DERIVATIVE FUNCTION,
TO FIND THE FINAL DERIVATIVE FUNCTION,

AND THAT WOULD BE 24X SQUARED + 12X + 11.
AND THAT WOULD BE 24X SQUARED + 12X + 11.

THIS IS A DERIVATIVE OF THE INITIAL FUNCTION.
THIS IS A DERIVATIVE OF THE INITIAL FUNCTION.

NOW YOU MIGHT ASK YOURSELF,
NOW YOU MIGHT ASK YOURSELF,

COULD YOU JUST MULTIPLY ALL OF THIS OUT
COULD YOU JUST MULTIPLY ALL OF THIS OUT

AND THEN FIND THE DERIVATIVE JUST USING THE BASIC POWER RULE,
AND THEN FIND THE DERIVATIVE JUST USING THE BASIC POWER RULE,

AND IN FACT YOU COULD.
AND IN FACT YOU COULD.

I'M NOT GOING TO TAKE TIME TO DO IT, BUT I WILL SHOW IT.
I'M NOT GOING TO TAKE TIME TO DO IT, BUT I WILL SHOW IT.

IF YOU WERE TO TAKE THE TIME TO MULTIPLY ALL OF THIS OUT,
IF YOU WERE TO TAKE THE TIME TO MULTIPLY ALL OF THIS OUT,

THERE WOULD BE SIX DIFFERENT PRODUCTS.
THERE WOULD BE SIX DIFFERENT PRODUCTS.

YOU WOULD GET THIS.
YOU WOULD GET THIS.

THEN IF YOU COMBINED YOUR LIKE TERMS, YOU WOULD HAVE THIS.
THEN IF YOU COMBINED YOUR LIKE TERMS, YOU WOULD HAVE THIS.

AND THEN IF YOU JUST FOUND THE DERIVATIVE OF EACH TERM,
AND THEN IF YOU JUST FOUND THE DERIVATIVE OF EACH TERM,

YOU CAN SEE THAT THIS DERIVATIVE MATCHES THE DERIVATIVE
YOU CAN SEE THAT THIS DERIVATIVE MATCHES THE DERIVATIVE

FROM THE PREVIOUS SLIDE, OF COURSE IT'S THE SAME FUNCTION,
FROM THE PREVIOUS SLIDE, OF COURSE IT'S THE SAME FUNCTION,

SO EITHER WAY, IT WORKS.
SO EITHER WAY, IT WORKS.

BUT WE DO WANT TO USE THE PRODUCT RULE IN MANY CASES,
BUT WE DO WANT TO USE THE PRODUCT RULE IN MANY CASES,

BECAUSE IT IS GOING TO BE A LOT EASIER IN MANY CASES.
BECAUSE IT IS GOING TO BE A LOT EASIER IN MANY CASES.

LET'S TAKE A LOOK AT ANOTHER.
LET'S TAKE A LOOK AT ANOTHER.

HERE'S A PROBLEM WHERE IT MIGHT BE A LOT OF WORK
HERE'S A PROBLEM WHERE IT MIGHT BE A LOT OF WORK

TO MULTIPLY ALL THIS OUT.
TO MULTIPLY ALL THIS OUT.

LET'S TRY TO FIND THE DERIVATIVE USING THE PRODUCT RULE.
LET'S TRY TO FIND THE DERIVATIVE USING THE PRODUCT RULE.

I'M FIRST GOING TO WRITE UP WHAT WE'RE GOING TO DO,
I'M FIRST GOING TO WRITE UP WHAT WE'RE GOING TO DO,

AND THEN WE'LL DO IT.
AND THEN WE'LL DO IT.

OKAY, SO AGAIN, THIS WOULD BE OUR FIRST FUNCTION.
OKAY, SO AGAIN, THIS WOULD BE OUR FIRST FUNCTION.

THIS WOULD BE OUR SECOND FUNCTION.
THIS WOULD BE OUR SECOND FUNCTION.

HERE'S OUR PRODUCT RULE FOR REFERENCE.
HERE'S OUR PRODUCT RULE FOR REFERENCE.

WE HAVE F x THE DERIVATIVE OF G + G x THE DERIVATIVE OF F.
WE HAVE F x THE DERIVATIVE OF G + G x THE DERIVATIVE OF F.

OUR NEXT STEP WILL BE TO FIND THESE DERIVATIVES
OUR NEXT STEP WILL BE TO FIND THESE DERIVATIVES

AND THEN MULTIPLY AND SIMPLIFY.
AND THEN MULTIPLY AND SIMPLIFY.

THE DERIVATIVE OF X SQUARED + 3X - 9 WOULD BE 2X + 3.
THE DERIVATIVE OF X SQUARED + 3X - 9 WOULD BE 2X + 3.

AND THE DERIVATIVE OF 3X SQUARED - X - 5 WOULD BE 6X - 1.
AND THE DERIVATIVE OF 3X SQUARED - X - 5 WOULD BE 6X - 1.

SO IF WE MULTIPLY ALL THIS OUT,
SO IF WE MULTIPLY ALL THIS OUT,

YOU CAN SEE THERE'S GOING TO BE A LOT OF DIFFERENT PRODUCTS.
YOU CAN SEE THERE'S GOING TO BE A LOT OF DIFFERENT PRODUCTS.

YOU MIGHT WANT TO PAUSE THE VIDEO AND DOUBLE-CHECK THIS.
YOU MIGHT WANT TO PAUSE THE VIDEO AND DOUBLE-CHECK THIS.

BUT NOW, WE HAVE TO COMBINE OUR LIKE TERMS,
BUT NOW, WE HAVE TO COMBINE OUR LIKE TERMS,

SO LET'S FIRST IDENTIFY THEM.
SO LET'S FIRST IDENTIFY THEM.

SO THIS FINALLY SIMPLIFIES TO
SO THIS FINALLY SIMPLIFIES TO

-- CUBE + 24X SQUARED - 50X + 24.
-- CUBE + 24X SQUARED - 50X + 24.

SO AS YOU CAN SEE, THERE'S A LOT OF ALGEBRA
SO AS YOU CAN SEE, THERE'S A LOT OF ALGEBRA

IN FINDING THESE DERIVATIVES.
IN FINDING THESE DERIVATIVES.

AND OF COURSE IF YOU MISS ONE SIGN,
AND OF COURSE IF YOU MISS ONE SIGN,

IT'S GOING TO THROW EVERYTHING OFF
IT'S GOING TO THROW EVERYTHING OFF

AND IT'S GOING TO BE VERY DIFFICULT TO GO BACK
AND IT'S GOING TO BE VERY DIFFICULT TO GO BACK

AND FIND YOUR ERROR.
AND FIND YOUR ERROR.

SO YOU HAVE TO BE EXTREMELY CAREFUL
SO YOU HAVE TO BE EXTREMELY CAREFUL

AND IF YOU DO SO, YOU'LL BE REWARDED.
AND IF YOU DO SO, YOU'LL BE REWARDED.

LET'S TAKE A LOOK AT ONE LAST PROBLEM.
LET'S TAKE A LOOK AT ONE LAST PROBLEM.

GIVEN THE FUNCTION F OF X,
GIVEN THE FUNCTION F OF X,

FIND THE EQUATION OF THE TANGENT LINE AT X = -2,
FIND THE EQUATION OF THE TANGENT LINE AT X = -2,

AND THERE'S A LOT GOING ON IN THIS PROBLEM.
AND THERE'S A LOT GOING ON IN THIS PROBLEM.

THE FIRST THING I WANT TO DO IS TAKE A LOOK
THE FIRST THING I WANT TO DO IS TAKE A LOOK

AT THE GRAPH OF THIS FUNCTION.
AT THE GRAPH OF THIS FUNCTION.

SO WE'RE GOING TO TYPE THE FUNCTION INTO Y1.
SO WE'RE GOING TO TYPE THE FUNCTION INTO Y1.

LET'S GO AHEAD AND ADJUST THE WINDOW TO THE FOLLOWING,
LET'S GO AHEAD AND ADJUST THE WINDOW TO THE FOLLOWING,

-3 TO 3 ON THE X-AXIS AND LET'S GO FROM -100 TO 100 ON Y-AXIS.
-3 TO 3 ON THE X-AXIS AND LET'S GO FROM -100 TO 100 ON Y-AXIS.

THAT'S A GRAPH.
THAT'S A GRAPH.

OKAY, SO HERE'S OUR FUNCTION.
OKAY, SO HERE'S OUR FUNCTION.

THE FIRST THING WE NEED TO RECOGNIZE
THE FIRST THING WE NEED TO RECOGNIZE

IS IN ORDER TO FIND A TANGENT LINE,
IS IN ORDER TO FIND A TANGENT LINE,

WE'RE GOING TO HAVE TO FIND THE POINT ON THE FUNCTION,
WE'RE GOING TO HAVE TO FIND THE POINT ON THE FUNCTION,

NOT JUST WHERE X = -2.
NOT JUST WHERE X = -2.

AN EASY WAY TO DO THAT IS TO HIT SECOND GRAPH AND TYPE IN -2
AN EASY WAY TO DO THAT IS TO HIT SECOND GRAPH AND TYPE IN -2

IF YOU HAVE IT ON THE ASKED FEATURE,
IF YOU HAVE IT ON THE ASKED FEATURE,

OR YOU CAN SCROLL AROUND AND FIND THE -2.
OR YOU CAN SCROLL AROUND AND FIND THE -2.

SO THE POINT OF TANGENCY IS GOING TO BE (-2,-3).
SO THE POINT OF TANGENCY IS GOING TO BE (-2,-3).

LET'S WRITE THAT DOWN, OUR POINT WILL BE (-2,-3).
LET'S WRITE THAT DOWN, OUR POINT WILL BE (-2,-3).

NOW IN ORDER TO FIND THE EQUATION OF A LINE,
NOW IN ORDER TO FIND THE EQUATION OF A LINE,

WE HAVE TO FIND THE SLOPE OF THE TANGENT LINE.
WE HAVE TO FIND THE SLOPE OF THE TANGENT LINE.

IN ORDER TO FIND THE SLOPE OF THE TANGENT LINE,
IN ORDER TO FIND THE SLOPE OF THE TANGENT LINE,

WE HAVE TO FIND THE DERIVATIVE OF THIS FUNCTION.
WE HAVE TO FIND THE DERIVATIVE OF THIS FUNCTION.

SO LET'S GO AHEAD AND DO THAT.
SO LET'S GO AHEAD AND DO THAT.

WE'RE GOING TO HAVE TO USE OUR PRODUCT RULE.
WE'RE GOING TO HAVE TO USE OUR PRODUCT RULE.

THERE'S OUR FIRST FUNCTION.
THERE'S OUR FIRST FUNCTION.

THERE'S OUR SECOND FUNCTION.
THERE'S OUR SECOND FUNCTION.

LET'S WRITE OUT THE PRODUCT RULE,
LET'S WRITE OUT THE PRODUCT RULE,

AND THEN WE'LL FIND THE DERIVATIVE.
AND THEN WE'LL FIND THE DERIVATIVE.

OKAY, SO I HAVE APPLIED PRODUCT RULE.
OKAY, SO I HAVE APPLIED PRODUCT RULE.

ONE THING I WANT YOU TO NOTICE IS I REMEMBERED
ONE THING I WANT YOU TO NOTICE IS I REMEMBERED

THAT IN ORDER TO FIND THE DERIVATIVE OF 2 DIVIDED BY X,
THAT IN ORDER TO FIND THE DERIVATIVE OF 2 DIVIDED BY X,

WE HAVE TO REWRITE THIS AS 2X TO THE POWER OF -1.
WE HAVE TO REWRITE THIS AS 2X TO THE POWER OF -1.

SO I DID THAT HERE FOR F
SO I DID THAT HERE FOR F

AND OVER HERE WHEN WE FIND THE DERIVATIVE OF F.
AND OVER HERE WHEN WE FIND THE DERIVATIVE OF F.

LET'S GO AHEAD AND FIND THE DERIVATIVES AND THEN SIMPLIFY.
LET'S GO AHEAD AND FIND THE DERIVATIVES AND THEN SIMPLIFY.

THE DERIVATIVE OF X SQUARED - 3 WOULD BE 2X.
THE DERIVATIVE OF X SQUARED - 3 WOULD BE 2X.

THE DERIVATIVE OF X + 2X TO THE -1 POWER
THE DERIVATIVE OF X + 2X TO THE -1 POWER

WOULD BE 1 - 2X TO THE POWER OF -2.
WOULD BE 1 - 2X TO THE POWER OF -2.

SO NOW, WE HAVE TO MULTIPLY THIS OUT AND SIMPLIFY.
SO NOW, WE HAVE TO MULTIPLY THIS OUT AND SIMPLIFY.

2X x X WOULD BE 2X SQUARED.
2X x X WOULD BE 2X SQUARED.

NOW, 2X x 2X TO THE POWER OF -1 THAT WOULD ACTUALLY BE 4.
NOW, 2X x 2X TO THE POWER OF -1 THAT WOULD ACTUALLY BE 4.

REMEMBER WHEN YOU MULTIPLY X TO THE -1 x X,
REMEMBER WHEN YOU MULTIPLY X TO THE -1 x X,

YOU ADD YOUR EXPONENTS.
YOU ADD YOUR EXPONENTS.

-1 + 1 WOULD BE 0, X TO THE 0 IS 1,
-1 + 1 WOULD BE 0, X TO THE 0 IS 1,

+ X SQUARED (-2,-2).
+ X SQUARED (-2,-2).

AGAIN ON THIS PRODUCT, THE Xs SIMPLIFY TO 1 - 3,
AGAIN ON THIS PRODUCT, THE Xs SIMPLIFY TO 1 - 3,

AND HERE WE HAVE 6X TO THE POWER OF -2.
AND HERE WE HAVE 6X TO THE POWER OF -2.

SIMPLIFYING THIS, WE WOULD GET 3X SQUARED (-1,6)
SIMPLIFYING THIS, WE WOULD GET 3X SQUARED (-1,6)

DIVIDED BY X SQUARED.
DIVIDED BY X SQUARED.

NOW, I'VE GONE A LITTLE QUICKLY OVER SOME OF THE ALGEBRA
NOW, I'VE GONE A LITTLE QUICKLY OVER SOME OF THE ALGEBRA

JUST DUE TO THE TIME CONSTRAINT.
JUST DUE TO THE TIME CONSTRAINT.

YOU MIGHT WANT TO PAUSE THE VIDEO
YOU MIGHT WANT TO PAUSE THE VIDEO

AND TAKE A CLOSE LOOK IF YOU NEED TO.
AND TAKE A CLOSE LOOK IF YOU NEED TO.

NOW TO FIND THE SLOPE,
NOW TO FIND THE SLOPE,

WE NEED TO EVALUATE THE DERIVATIVE AT X = -2.
WE NEED TO EVALUATE THE DERIVATIVE AT X = -2.

THIS WOULD GIVE US 12.5,
THIS WOULD GIVE US 12.5,

THAT WOULD BE THE SLOPE OF THE TANGENT LINE AT X = -2.
THAT WOULD BE THE SLOPE OF THE TANGENT LINE AT X = -2.

LET'S GO AHEAD AND FIND THE EQUATION NOW.
LET'S GO AHEAD AND FIND THE EQUATION NOW.

RECALL THAT OUR POINT WAS (-2,- 3) AND OUR SLOPE IS 12.5.
RECALL THAT OUR POINT WAS (-2,- 3) AND OUR SLOPE IS 12.5.

SO USING POINT SLOPE FORMULA,
SO USING POINT SLOPE FORMULA,

WE'D HAVE THIS WHICH WE'D SIMPLIFY TO Y + 3
WE'D HAVE THIS WHICH WE'D SIMPLIFY TO Y + 3

IS EQUAL TO 12.5 x X + 2.
IS EQUAL TO 12.5 x X + 2.

SO DISTRIBUTE HERE AND SOLVE FOR Y -- 3 ON BOTH SIDES,
SO DISTRIBUTE HERE AND SOLVE FOR Y -- 3 ON BOTH SIDES,

WE WOULD OBTAIN THE EQUATION OF THE TANGENT LINE.
WE WOULD OBTAIN THE EQUATION OF THE TANGENT LINE.

GO AHEAD AND VERIFY THAT.
GO AHEAD AND VERIFY THAT.

HERE'S OUR TANGENT LINE IN RED,
HERE'S OUR TANGENT LINE IN RED,

AND IT DOES LOOK LIKE ITS TANGENT AT X = -2.
AND IT DOES LOOK LIKE ITS TANGENT AT X = -2.

I'D LIKE TO GO BACK TO THE GRAPHING CALCULATOR
I'D LIKE TO GO BACK TO THE GRAPHING CALCULATOR

FOR A MOMENT.
FOR A MOMENT.

SO IF WE GO BACK TO OUR GRAPH,
SO IF WE GO BACK TO OUR GRAPH,

RIGHT ABOVE THE TRACE BUTTON, THERE'S A CALC MENU.
RIGHT ABOVE THE TRACE BUTTON, THERE'S A CALC MENU.

IF I HIT SECOND TRACE,
IF I HIT SECOND TRACE,

OPTION 6 WILL ALLOW YOU TO FIND THE VALUE
OPTION 6 WILL ALLOW YOU TO FIND THE VALUE

OF THE DERIVATIVE AT A GIVEN X VALUE.
OF THE DERIVATIVE AT A GIVEN X VALUE.

SO IF I SELECT OPTION 6 AND THEN TYPE IN -2
SO IF I SELECT OPTION 6 AND THEN TYPE IN -2

WHERE WE WANT TO FIND THE DERIVATIVE,
WHERE WE WANT TO FIND THE DERIVATIVE,

IT WILL VERIFY OUR SLOPE OF THE TANGENT LINE
IT WILL VERIFY OUR SLOPE OF THE TANGENT LINE

TO BE 12.5 AT X = -2.
TO BE 12.5 AT X = -2.

I HOPE THAT HELPS YOU FIND DERIVATIVES
I HOPE THAT HELPS YOU FIND DERIVATIVES

USING THE PRODUCT RULE.
USING THE PRODUCT RULE.

THANK YOU FOR WATCHING.
THANK YOU FOR WATCHING.