SHORT NOTES MODULE 2 : APPLICATIONS OF DERIVATIVE
TANGENT AND NORMAL LINE
y
(x1 , y1)
y = f(x)
Equation of tangent line:
x Equation of normal line:
RATE OF CHANGE
USING IMPLICIT DIFFERENTIATION
USING CHAIN RULE
example: 4 V = r 3 3
given :
dV dr
example: h=x+y
and
dV dV dr = dt dr dt
dr dt
dh dx dy = + dt dt dt
CURVE SKETCHING
② find x0 such that f '(x0)=0
① form intervals based on stationary number
① substitute test value into f '(x); ie f '(k)
CONCLUSION
① find f '(x)
TEST VALUES
INCREASING/DECREASING INTERVAL
INTERVALS
STATIONARY POINT
(I)
② take a test
① interval is increasing if f '(k) > 0 ② interval is decreasing if
value k within
f '(k) < 0
each interval
(II)
MAXIMUM/MINIMUM POINT
FIRST DERIVATIVE TEST
MAXIMUM POINT
f(x) changes from increasing to decreasing
(x, f(x)) is a maximum point
NO MAXIMUM/ MINIMUM POINT
MINIMUM POINT
f(x) changes from decreasing to increasing
(x, f(x)) is a minimum point
no changes to f(x)
f(x) remained increasing or decreasing
SECOND DERIVATIVE TEST
MAXIMUM POINT
MINIMUM POINT
f ‘’(x0) < 0
f ‘’(x0) > 0
NO CONCLUSION
f ‘’(x0) = 0
Apply 1st Derivative Test
(III)
CONCAVE UPWARD/DOWNWARD
ZERO OF f ''(x)
INTERVAL(S)
• find x0 such that f ''(x0) = 0
(IV)
• form interval(s) based on x0 • take a test value k within each interval
• substitute test value into f ''(x); ie f ''(k)
CONCLUSION
• f(x) is concave upward if f ''(k) > 0 • f(x) is concave downward if f ''(k) < 0
INFLECTION POINT
f(x) changes concavity at x0
INFLECTION POINT
(V)
TEST VALUES
GRAPH
y-intercept
x,y - axes
inflection point
max/min point(s) correct shape
correct position
(x0 , f(x0)) is an inflection point