.Application of derivatives Basic concept Class-12
(इस chapter के अंदर हम निम्नलिखित टॉपिक पढ़ने वाले हैं)
1. Rate of change of quantities
2. Increasing and decreasing functions
3. Tangent and Normal
4. Approximations
5. Maxima and Minima
| Exercise | Topic
|
| 6.1 | Introduction |
| 6.2 | Rate of Change of Quantities |
| 6.3 | Increasing and Decreasing Functions |
| 6.4 | Tangents and Normals |
| 6.5 | Approximations |
| 6.6 | Maxima and Minima |
| Others | Miscellaneous Q&A |
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1. Rate of change of quantities:
⬇️
⬇️
Area/Volume/C.S.A/T.S.A/Radius/Length / Breath. etc..
⬇️
⬇️
Variable---------------------------- ---Constant
↙️ ↘️.
1. Dependent. 1. Arbitrary
And ( ax+by+c=0)
2. Independent. 2. Absolute
( 2x+3y=8)
If the change in one variable y with respect to x then dy/dx = f'(x) denoted the rate of change with respect to x
lim x➡️0 ∆y/∆x = dy/dx
A = π r²
↙️. ↘️
Dep Independent
(बनने वाला) (बनाने वाला)
C = 2π r
↙️. ↘️
Dep Independent
(बनने वाला) (बनाने वाला)
(r में change होने से circle के Area और circumference दोनों change होंगे )
------------------------------------------------------------ A. ➡️r. ➡️ t
dA/dr × dr/dt = dA/dt
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2. Increasing and decreasing. functions.
(जब x का मान बढ़ाने पर y का मान भी बढ़ता हो तो function, Increasing होगा, और जब x का मान बढ़ाने पर y का मान घटता हो तो function decreasing होगा )
कोई function Increasing है या फिर decreasing है यह पता करने के निम्नलिखित दो तरीके हैं l
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1 . By graph. 2. By f'(x)
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1 . By graph.
Increasing function. ⬇️
⬇️
⬇️
Neither Increasing nor decreasing function.
⬇️
.
(A) x ≥ 0
(B) x ≤ 0
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(ii) y = 7
Neither Increasing nor Decreasing
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(iii) y = 2x + 1. Increasing
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(iv) y = -3x + 2. decreasin
----------------------------------------------------------- 2. By f'(x)-------
CONCEPT
(A). f is Increasing in [a b] if f'(x) ≥ 0 for each X ∈ [a b ]
(B). f is Decreasing in [a b] if f'(x) ≤ 0 for each X ∈ [a b ]
(C). f is contant [a b ] if f'(x) = 0 for X ∈ [a b ]
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WORKING RULE FOR INCREASING AND DECREASING
(i) Function
(ii) Differentiation
(iii) Linear Factor
(iv) f'(x) = 0
(v) Find the values of x (turning point/Extreme point)
(vi) Show the value of x on number line
(vii) Write Intervals
(viii) Find f'(x) at any simple x in Interval
f'(x) ≥ 0 , Increasing
f'(x) > 0, Strictly Increasing
f'(x) ≤ 0, Decreasing
f'(x) < 0, Strictly Decreasing
Example f(x) = x² - x - 6
f'(x) = 2x -1
f'(x) = 0
2x -1 = 0
x = 1/2
- ∞ -------------------- 1/2 ------------- ∞.
Intervals | Sign of f’(x) | Nature |
| ( -∞ 1/2 ) | - ( negative) | Strictly Decreasing |
| ( 1/2 ∞) | + ( Positive) | Strictly Increasing |
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(Out of syllabus)
3. Tangent and Normal.
Equation of tangent. y-y1 = dy/dx( x-x1)
Equation of Normal. y-y1 = - dx/dy( x-x1)
When tangent is parallel to x-axis then:
dy/dx =0
When tangent is perpendicular to x-axis then:
dx/dy = 0
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(Out of syllabu
4.Approximations
f (x+∆x ) = f(x) + ∆x.f'(x)
Ex: √(25.1) = √ (25+ 0.1)
↙️ ↘️
x ∆x
f(x) = √x f'(x) = 1/2√x
= f(x) + ∆x. f'(x)
= √x + ∆x• 1/2√x
= √25 + 0.1 × 1/2√25
= 5. +0.1/10
=. 5 + 0.01
=. 5.01. Ans
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5. Maxima and Minima.
Application of derivatives (A.O.D) का सबसे Important topics है ,नाम से ही पता लग रहा है कि किसके विषय में बात होने वाली है l
अधिकतम और न्यूनतम
Maxima and Minima का पता हम निम्नलिखित तीन तरीकों से लगा सकते हैंl
1. Graph देख कर ,
2. By First derivatives test.
3. By Second derivatives test
1. Graph देख कर ,:. ........
जिस domain पर function defined है उसके अंदर सबसे छोटी value को function की minimum तथा सबसे बड़ी value को function की maximum value कहते हैं l
(A)
(B)
(C) (D)
(E)
(L)
एक function बहुत सारी maxima या minima values सकती हैं l इन सभी को Local Maxima या Local minima जाता है इन सभी में जो सबसे बड़ा होगा उसे Absolute Maxima तथा जो सबसे छोटा होगा उसे Absolute Minima जाता है l
Absolute Maxima ⊂ Local Maxima
Absolute Minima ⊂ Local Minima
(M)
()
2. By First derivatives test.
(i) Function
(ii) Differentiation
(iii) Linear Factor
(iv) f'(x) = 0
(v) Find the values of x (turning point/Extreme point/ critical point)
f'(x) का sign: + ➡️ - ( maximum)
- ➡️+ ( Minimum)
Graph(D)
Example : f(x) = x² - x - 6
f'(x) = 2x -1
at f'(x) = 0
2x -1 = 0
x = 1/2
-∞. -------------------- 1/2 ------------- ∞.
At value | Sign of f’(x) |
|
| ( 1/2- h) | - ( negative) | (- to +) MINIMUM |
| ( 1/2+h) | + ( Positive) |
|
Example: f(x) = x³ - 6x² + 9x + 15
f'(x)= 3x² - 12x + 9
= 3(x-1) (x-3)
at f'(x) =0 , x = 1 ,3
-------------------⬇️--------------------------⬇️-------------- f'(x)=.
(+) ⬇️.(-) (-) ⬇️(+)
(1-h)----•1•---(1+h)-----(3-h)----•3•------(3+h)
at x=1,f(x) will be maximum.
Maximum vale= f(1)=19
at x=3, f(x) will be minimum
Minimum value=f((3)= 15
Note: When the sign of f'(x) doesn't change then it's called point of inflation
Ex: f(x)= x³
f'(x)= 3x²
at f'(x) =0. x=0
--------------- ⬇️---------------------------------------
f'(x)=. (+) ⬇️.(+) (No change)
(0-h)----•0•---(0+h)----
f'(0-h)= (+)
f'(0+h)= (+) so x=0 is point of inflation
3. By Second derivatives test.
i) Function
(ii) Differentiation
(iii) Linear Factor
(iv) f'(x) = 0
(v) Find the values of x =a,b,c..
(turning point/Extreme point/ critical point)
(vi) f''(x) at a,b,c ..
(vii) If f''(x) = (+) then f(x) is Minimum
If f''(x) = (-) then f(x) is Maximum
f''(x)= 0 ( go to first derivatives test)
EXAMPLE: f(x) = 3x⁴ + 4x³- 12x² + 12
f'(x) = 12x³ + 12x² - 24x
= 12x( x+2) (x -1)
at f'(x) = 0 , x = 0 ,1 , -2
f''(x) = 36x² + 24x - 24
at x=0, f"(x)= -24(Neg) [ MAXIMUM]
⬇️
Point of Maxima
at x=1, f"(x)= 36 (Positive) [ MINIMUM]
⬇️
Point of Minima
at x= -2, f"(x)= 72 (Positive) [MINIMUM]
⬇️
Point of Minima
f(0) =12
f(1)= 7
f(-2)= -20
. MODULUS FUNCTION AND TRIGONOMETRY FUNCTION .
(i).
f(x)= |x+2| -1↗️ ((x+2)-1
↘️-(x+2)-1
f'(x)= ↗️ 1
↘️-1.
(Critical point not possible )
Then
|x+2| ≥ 0
|x+2| -1 ≥ -1
f(x) ≥ -1
So -1 is minimum value .
for point |x+2| -1 = 0
x - 2 = 0
x= -2 (point of Minima)
(ii).
g(x) = - |x+1| + 3
we know that. :|x+1| ≥ 0
- |x+1|≤ 0 ,
( Mode or Neg mei sign change krte hai)
- |x+1| + 3 ≤ 0 + 3
g(x) ≤ 3 (Maximum)
for point |x+1| = 0
x +1 = 0
x= -1 (point of Maxima)
(iii)
h(x) = sin2x + 5
-1 ≤ sin2x ≤ 1
-1 +5 ≤ sin2x ≤ 1+5
4 ≤ h(x). ≤ 6
Mini. Maxi
Point sin2x = 0. x ,=nπ +(-1)ñalpha
®️®️®️®️®️®️®️®️®️ ®️®️
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