short notes module 2 app of derv

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