- IF COSINE X = 2 SINE X,
WE WANT TO FIND F PRIME OF X 
AND F PRIME OF PI/4.
SINCE THE DERIVATIVE OF SINE X, 
WITH RESPECT TO X,
IS EQUAL TO COSINE X,
THE DERIVATIVE OF 2 SINE X 
WOULD BE 2 COSINE X.
SO TO FIND F PRIME OF PI/4
WE'LL SUBSTITUTE PI/4 FOR X 
IN THE DERIVATIVE FUNCTION.
THIS WOULD GIVE US 
2 COSINE OF PI DIVIDED BY 4.
WE CAN FIND THIS COSINE FUNCTION 
VALUE TWO WAYS.
WE COULD USE 
A REFERENCE TRIANGLE,
RECOGNIZING THIS 
IS A 45 DEGREE ANGLE,
OR WE COULD USE THE UNIT CIRCLE.
LET'S GO AHEAD 
AND SHOW BOTH WAYS.
IF THIS IS PI/4 RADIANS WE HAVE 
A 45-45-90 RIGHT TRIANGLE.
SO WE CAN LABEL 
THE TWO LEGS WITH LENGTH 1
AND THE HYPOTENUSE 
LENGTH SQUARE ROOT 2.
SO COSINE PI/4 
WOULD BE THE RATIO
OF THE ADJACENT SIDE 
TO THE HYPOTENUSE,
WHICH WOULD BE 1 
DIVIDED BY SQUARE ROOT 2.
OR IF WE RATIONALIZE THIS,
IT WOULD BE SQUARE ROOT 2 
DIVIDED BY 2.
OR USING THE UNIT CIRCLE,
HERE'S THE TERMINAL SIDE 
OF PI/4 RADIANS.
ON THE UNIT CIRCLE X = COSINE 
THETA AND Y = SINE THETA.
NOTICE HOW THE X COORDINATE HERE 
IS SQUARE ROOT 2 DIVIDED BY 2.
THEREFORE, 2 x COSINE PI/4 
WOULD BE EQUAL TO 2
OR 2/1 x SQUARE ROOT 2 
DIVIDED BY 2.
NOTICE HOW THE 2 SIMPLIFY TO 1 
HERE
LEAVING US WITH A DERIVATIVE 
FUNCTION VALUE OF SQUARE ROOT 2.
AS A DECIMAL THIS WOULD BE 
APPROXIMATELY 1.414,
WHICH MEANS IF WE WERE TO SKETCH 
THE ORIGINAL FUNCTION,
F OF X = 2 SINE X,
AND LOCATE THE POINT 
ON THE FUNCTION WHERE X = PI/4,
THE TANGENT LINE WOULD HAVE 
A SLOPE OF SQUARE ROOT 2
OR APPROXIMATELY 1.414.
LET'S GO AND TAKE A LOOK 
AT THAT.
AGAIN, HERE'S A GRAPH 
OF F OF X = 2 SINE X.
HERE'S WHERE X = PI/4 RADIANS.
LOOKING AT THIS POINT 
ON THE FUNCTION,
IF WE SKETCH THE TANGENT LINE 
AT THAT POINT,
THE SLOPE OF THIS TANGENT LINE 
WOULD BE SQUARE ROOT 2
OR APPROXIMATELY 1.414.
SO WHEN WE EVALUATE 
A DERIVATIVE FUNCTION
AT A SPECIFIC VALUE OF X,
KEEP IN MIND WE'RE DETERMINING 
THE SLOPE OF A TANGENT LINE
AT THAT POINT ON THE FUNCTION.
LET'S TAKE A LOOK 
AT OUR SECOND EXAMPLE.
WE HAVE G OF X = -2 COSINE X.
WE WANT TO FIND G PRIME OF X 
AND G PRIME OF 7PI/6.
NOTICE THE DERIVATIVE OF COSINE 
X IS EQUAL TO NEGATIVE SINE X.
SO G PRIME OF X WOULD BE 
-2 x -SINE X,
WHICH WOULD BE 2 SINE X.
SO TO FIND G PRIME 7PI/6
WE'LL NOW SUBSTITUTE 7PI/6 FOR X 
IN OUR DERIVATIVE FUNCTION.
THAT WOULD GIVE US 2 
x SINE 7PI DIVIDED BY 6.
LET'S GO AHEAD AND USE 
THE UNIT CIRCLE
TO DETERMINE 
THIS FUNCTION VALUE.
HERE'S A TERMINAL SIDE 
OF THE ANGLE 7PI/6 RADIANS.
REMEMBER SINE THETA IS EQUAL TO 
Y ON THE UNIT CIRCLE.
SO SINE 7PI/6 IS EQUAL TO -1/2.
SO WE'D HAVE 2 x -1/2 
OR IF WE WANT 2/1.
AND NOTICE HOW, AGAIN, THIS 
SIMPLIFIES NICELY TO JUST -1,
WHICH MEANS IF WE SKETCH THE 
FUNCTION G OF X = -2 COSINE X,
AND LOCATE THE POINT 
WHEN X = 7PI/6 RADIANS,
THE SLOPE OF THE TANGENT LINE 
AT THAT POINT WOULD BE -1.
LET'S TAKE A LOOK.
HERE'S A GRAPH OF THE ORIGINAL 
FUNCTION G OF X,
HERE'S WHERE X = 7PI/6 RADIANS.
HERE'S THE CORRESPONDING POINT 
ON THE FUNCTION,
AND HERE'S THE TANGENT LINE.
THE SLOPE OF THIS TANGENT LINE 
IS -1.
I HOPE YOU FOUND THIS HELPFUL.
