- WELCOME TO A VIDEO 
ON THE DERIVATIVE
OF SINE AND COSINE FUNCTIONS.
THE DERIVATIVE OF SINE X 
IS EQUAL TO COSINE X,
AND THE DERIVATIVE OF COSINE X 
IS EQUAL TO NEGATIVE SINE X.
LET'S TAKE A LOOK AT A PROOF 
OF THIS FIRST DERIVATIVE.
IF WE APPLY THE LIMIT 
DEFINITION OF THE DERIVATIVE
WE WOULD HAVE THE LIMIT
AS H APPROACHES ZERO 
OF THIS QUOTIENT.
NOW, IF WE USED 
THE SUM IDENTITY FOR SINE,
AS WE SEE HERE,
WE COULD EXPAND THIS TERM 
FOR THE SUM OF THESE PRODUCTS.
NEXT, IF WE TAKE THIS 
FIRST TERM AND THIS TERM,
AND FACTOR OUT SINE X,
WE WOULD HAVE COSINE H - 1 
DIVIDED BY H.
THEN IF WE TOOK THE SECOND 
AND FACTORED OUT COSINE X
WE'D BE LEFT SINE H 
DIVIDED BY H.
NEXT, WHEN WE FIND THIS LIMIT,
WE END UP GETTING THOSE 
TWO SPECIAL LIMITS
THAT WE JUST DISCUSSED 
IN A PREVIOUS VIDEO
WHERE WE KNOW THAT THIS LIMIT 
IS EQUAL TO ZERO,
AND THIS LIMIT 
IS EQUAL TO ONE.
SO WE HAVE NEGATIVE SINE X x 0 
+ COSINE X x 1,
WHICH GIVES US A DERIVATIVE 
OF SINE X AS COSINE X.
LET'S TAKE A LOOK AT THIS 
GRAPHICALLY AS WELL.
HERE WE HAVE THE GRAPH 
OF SINE X
AND THEN WE HAVE 
THIS BLUE TANGENT LINE.
WHAT WE'RE GOING TO DO,
IS PLOT THE SLOPES OF THESE 
TANGENT LINES OVER HERE
TO FORM THE GRAPH 
OF THE DERIVATIVE OF SINE X.
SO IT LOOKS LIKE 
HERE OUR SLOPE
IS EQUAL TO APPROXIMATELY ONE.
THE TOP OF THIS 
RELATIVE MAXIMUM
WE HAVE A SLOPE OF ZERO.
NOTICE HOW WE PLOTTED A POINT 
ON THE X-AXIS HERE AND SO-ON.
SO, AGAIN, WE'RE PLOTTING THE 
SLOPES OF THE TANGENT LINES
ON THE RIGHT TO FORM 
THE DERIVATIVE FUNCTION.
AND WE CAN SEE THE DERIVATIVE 
OF THE GRAPH OF SINE X
FORMS THE FUNCTION F PRIME 
OF X = COSINE X.
SO THE DERIVATIVE 
OF THE SINE FUNCTION
IS EQUAL TO THE COSINE 
FUNCTION.
LET'S DO THE SAME 
FOR COSINE X.
SO, AGAIN, WE'RE GOING 
TO GRAPH THE DERIVATIVE
ON THE RIGHT
OR THE SLOPES 
OF THE TANGENT LINES
TO FORM THE DERIVATIVE 
FUNCTION OF Y = COSINE X.
AS YOU CAN SEE, 
THE SLOPES ARE NEGATIVE
SO WE'RE PLOTTING POINTS 
BELOW THE X-AXIS.
IT'S TURNING, 
THE SLOPE IS NOW ZERO
AND IT'S TURNING POSITIVE, 
BACK TO ZERO,
TURNING NEGATIVE, BACK TO ZERO 
AND POSITIVE AGAIN.
AND WHAT WE'LL NOTICE HERE,
THE DERIVATIVE OF THE COSINE 
FUNCTION GRAPHED HERE
IS ACTUALLY THE SINE FUNCTION 
REFLECTED ACROSS THE X-AXIS
OR F PRIME OF X 
IS EQUAL TO NEGATIVE SINE X.
NOW, OF COURSE THIS IS 
NOT A PROOF,
BUT YOU MAY WANT TO TRY 
TO PROVE THAT YOURSELF
USING A SIMILAR METHOD 
THAT WE DID
PROVING THE DERIVATIVE 
OF SINE X.
LET'S GO AHEAD AND APPLY 
THESE TWO DERIVATIVE FORMULAS
WITH SOME PROBLEMS.
WE WANT TO FIND THE DERIVATIVE 
OF EACH FUNCTION.
SO F PRIME OF X IS EQUAL 
TO THE DERIVATIVE OF X CUBE.
USING THE POWER RULE WE'D HAVE
3X SQUARED - 1/2 
x THE DERIVATIVE OF COSINE.
WELL, DERIVATIVE OF COSINE 
IS NEGATIVE SINE X.
SO WE NEED TO SIMPLIFY THIS.
SO WE'D HAVE 
3X SQUARED + 1/2 SINE X.
FOR THIS NEXT ONE 
WE CANNOT APPLY THE POWER RULE
IN THIS FORM.
LET'S GO AHEAD 
AND REWRITE THIS
AS X TO THE POWER 
OF -1 - 3 SINE X.
SO NOW WE CAN FIND G 
PRIME OF X.
THAT'LL BE EQUAL TO -1 x X TO 
THE -1 - 1 THAT'S -2 POWER - 3
x DERIVATIVE OF SINE X, 
WHICH IS COSINE X.
NOW, WE NEED TO CLEAN THIS UP 
A LITTLE BIT.
SO THIS WOULD BE -1/X SQUARED 
- 3 COSINE X.
LET'S TAKE A LOOK 
AT A FEW MORE EXAMPLES.
H PRIME OF X WILL BE EQUAL TO 
2 x THE DERIVATIVE OF SINE X,
WHICH IS COSINE X,
+ 3 TRANS DERIVATIVE OF COSINE 
X, WHICH IS NEGATIVE SINE X.
SO WE GOT TO WATCH THE SIGNS
WHEN WE FIND 
THESE DERIVATIVES.
LET'S GO AHEAD AND REWRITE 
THIS AS 2 COSINE X - 3 SINE X.
AND ONE MORE BASIC DERIVATIVE 
HERE,
LET'S REWRITE THIS 
AS X TO THE 1/2 POWER.
NOW WE'LL GO AHEAD 
AND APPLY THE POWER RULE
TO FIND THE DERIVATIVE 
OF THIS FIRST TERM.
SO WE'LL HAVE 4 x 1/2, 
X TO THE 1/2 - 1
THAT'D BE -1/2 + 3 
x THE DERIVATIVE OF COSINE X,
WHICH IS NEGATIVE SINE X.
LET'S GO AHEAD 
AND CLEAN THIS UP.
LOOKS LIKE WE'LL HAVE TWO.
AND THIS WOULD BECOME 
THE SQUARE ROOT OF X
IN THE DENOMINATOR.
MOVING THIS DOWN 
IT WOULD BE X TO THE 1/2,
WHICH IS THE SQUARE ROOT OF X.
AND THEN MINUS THE SINE X.
OKAY, LET'S GO AHEAD AND TAKE 
A LOOK AT ONE MORE PROBLEM
THAT'S A LITTLE MORE INVOLVED.
WE WANT TO DETERMINE 
THE POINTS
WHERE THE FUNCTION 
HAS HORIZONTAL TANGENT LINES
ON THE INTERVAL FROM 0 TO 2PI.
REMEMBER THE DERIVATIVE 
WOULD BE EQUAL TO ZERO
WHERE WE HAVE HORIZONTAL 
TANGENT LINES.
SO WE NEED TO FIND 
THE DERIVATIVE,
SET IT EQUAL TO ZERO, 
AND THEN SOLVE FOR X.
SO F PRIME OF X IS EQUAL TO--
THIS IS THE SQUARE ROOT OF 3 
x X TO THE FIRST.
SO THE DERIVATIVE OF X 
TO THE FIRST WOULD JUST BE 1.
SO WE HAVE SQUARE ROOT 3.
AND THE DERIVATIVE OF COSINE X 
IS NEGATIVE SINE X.
WE WANT TO KNOW 
WHERE THIS IS EQUAL TO ZERO,
SO WHAT WE'LL DO HERE 
IS CLEAN IT UP,
AND THEN SOLVE FOR SINE X.
ADD 2 SINE X TO BOTH SIDES, 
DIVIDE BY 2.
SO WE NEED TO FIND THE ANGLES
WHERE THE SINE FUNCTION VALUE 
IS EQUAL TO SQUARE ROOT 3/2.
THIS SHOULD REMIND YOU 
OF A 30-60-90 RIGHT TRIANGLE
WHERE A 60 DEGREE ANGLE 
HAS A SINE FUNCTION VALUE
OF SQUARE ROOT 3/2.
HOWEVER, THE SINE FUNCTION 
IS ALSO POSITIVE
IN THE SECOND QUADRANT,
SO WE NEED TO SKETCH 
A 60 DEGREE REFERENCE ANGLE
IN THE FIRST 
AND THE SECOND QUADRANT
TO DETERMINE BOTH ANGLES THAT 
HAVE A SINE FUNCTION VALUE
OF SQUARE ROOT 3/2.
SO X WILL EQUAL 60 DEGREES 
OR PI/3 RADIANS.
X COULD ALSO EQUAL 120 DEGREES 
OR 2PI DIVIDED BY 3 RADIANS.
NOW, IT'S ASKING 
FOR THE POINTS
WHERE THE FUNCTION 
HAS A HORIZONTAL TANGENT LINE.
SO WE ONLY FOUND 
THE X COORDINATE.
SO WE DO HAVE TO STILL GO BACK
AND FIND THE Y COORDINATES 
OF THOSE POINTS.
YEAH, LET'S TAKE 
THESE X VALUES,
SUB THEM BACK 
INTO THE ORIGINAL FUNCTION
TO FIND THE CORRESPONDING 
Y VALUES.
LET'S TAKE THIS INFORMATION 
OVER TO THE NEXT SCREEN.
SO WHEN X IS EQUAL TO PI/3,
F OF X WILL EQUAL SQUARE ROOT 
3 x PI/3 + 2 COSINE PI/3.
SO USING THE REFERENCE 
TRIANGLE
IN THE FIRST QUADRANT,
COSINE OF PI/3 
IS GOING TO EQUAL 1/2.
SO WE'LL HAVE SQUARE ROOT 3PI 
DIVIDED BY 3 + 2 x 1/2,
WHICH IS EQUAL TO 1.
AND 1 IS THE SAME AS 3/3.
SO WE'D HAVE THE SQUARE ROOT 
3PI + 3/3
FOR THE Y COORDINATE.
SO THAT POINT IS GOING TO BE 
(PI/3,SQUARE ROOT 3PI + 3/3).
NEXT, WHEN X IS EQUAL TO 2PI/3
WE'LL USE THE REFERENCE 
TRIANGLE
IN THE SECOND QUADRANT
TO FIND F OF X.
AND THE ONLY DIFFERENCE HERE
IS NOW THAT COSINE 2PI/3 
IS GOING TO BE -1/2.
SO WE'LL HAVE 2 SQUARE ROOT 
3PI/3 + 2 x -1/2.
SO INSTEAD OF HAVING A +3 HERE 
WE'LL HAVE A -3.
SO THE ORDERED PAIR 
WILL BE 2PI/3,
2 SQUARE ROOT 3PI/3 - 3.
AGAIN, THIS WOULD BE - 1 
WHICH IS THE SAME AS -3/3,
AS WE SEE HERE.
I HOPE YOU FOUND THIS VIDEO 
HELPFUL.
THANK YOU FOR WATCHING.
