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    Md. Aktar Kamal Assistant Professor (Management) FBS, Bup

    Uploaded by

    Aminul Islam Amu
    1. Differential calculus deals with calculating rates of change and finding derivatives. 2. The passage explains basic rules of differentiation, such as how to take the derivative of functions involving constants, sums, products, quotients, and powers. 3. Examples are provided to illustrate how to apply the rules to find derivatives of various functions.

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    Md. Aktar Kamal Assistant Professor (Management) FBS, Bup

    Uploaded by

    Aminul Islam Amu
    29 views20 pages
    1. Differential calculus deals with calculating rates of change and finding derivatives. 2. The passage explains basic rules of differentiation, such as how to take the derivative of functions involving constants, sums, products, quotients, and powers. 3. Examples are provided to illustrate how to apply the rules to find derivatives of various functions.

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    Lecture-5 - Differentiation.pptx

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    1. Differential calculus deals with calculating rates of change and finding derivatives. 2. The passage explains basic rules of differentiation, such as how to take the derivative of functions involving constants, sums, products, quotients, and powers. 3. Examples are provided to illustrate how to apply the rules to find derivatives of various functions.

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    29 views20 pages

    Md. Aktar Kamal Assistant Professor (Management) FBS, Bup

    Uploaded by

    Aminul Islam Amu
    1. Differential calculus deals with calculating rates of change and finding derivatives. 2. The passage explains basic rules of differentiation, such as how to take the derivative of functions involving constants, sums, products, quotients, and powers. 3. Examples are provided to illustrate how to apply the rules to find derivatives of various functions.

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    of 20

    DIFFERENTIAL

    CALCULUS

    Md. Aktar Kamal


    Assistant Professor ( Management)
    FBS,BUP
    DIFFERENTIAL CALCULUS

    CALCULUS gives us the technique for


    measuring changes in the dependent
    variable with the reference to a very small
    change, approaching almost zero, in the
    independent variables or variables
    DIFFERENTIAL CALCULUS
    The techniques concerning the calculation of
    the average rate of change are studies under
    differentiation or the Differential Calculus and
    the calculation of the total amount of change
    in the given range of values is studied under
    integration or integral Calculus.
    DIFFERENTIAL CALCULUS
    Let us assume that y has been produced by
    labor x and that as we increase x (labor) by
    one unit, the amount of y increases by four
    units. This relationship is shown by y = 4x
    DIFFERENTIAL CALCULUS
    When x is increased by a small increment Δx, then y
    increases by Δy, and
    we have y + Δy = 4(x +Δx) = 4x + 4Δx = y + 4Δx
    ⇒ Δy = 4Δx
    ⇒ Δy /Δx = 4
    Δy /Δx is the incremental ratio of dependent variable y
    with respect to the independent variable x , i.e. we can
    say Δy /Δx is the change in y with respect to a small unit
    change in x.
    Rules

    (1) (xn ) = n xn-1


    (2) (k) = 0 Where k is a constant, i.e. The derivative of a
    constant is 0.
    (3) [ kf(x)]= k [f(x)]= kf/(x) Where k is a constant.
    (4) [ f(x) ±g(x)] = f(x) ± g(x)
    (5) [ f(x) g(x)] = f(x) [g(x)] + g(x) [f(x)]
    (6) If y = u/v then = where v ≠ 0.
    (7) If y = x then (x) = 1
    Rules –
    Rules
    Rule – 1: ( xn ) = n xn-1
    Find the first derivative of the following function? if
    (a) f(x)= x10
    (b) f(x)= x 2/3 (c) f(x) =
    Solution: (a) Given f(x) = x10
    Differentiating w. r. t. ‘x’ then we have
    ∴ f(x) = (x10 ) = 10 x10-1 = 10 x9 ∴ f/(x)
    = 10 x9
    Rules
    Rules
    If g(x) = x find g /(x)
     
    Solution : Given g(x) = x6.5
    ∴ g(x) = ( x 6.5) = 6.5 X5.5
    ∴ g/(x) = 6.5 x 5.5
    Rules

    Rule – 2: If k is any constant then (k) = 0


    Example: If f(x) = 10 then f(x) = (10) = 0
    Rules
    Rule-3: [k f(x)] = k [f(x)]= k f/ (x)
    Where k is a constant.
    Example:
    (a) If y = 2x then find =?
    Solution:
    Given y = 2x
    Differentiating w. r. t. ‘x’ then we have
    ∴ (y) = (2x) = 2 (x) = 2 × 1 = 2
    ∴ y1 = 2
    Rules
    (b) If y = 3x2 then find =?
    Solution:
    Given y = 3x2
    Differentiating w. r. t. ‘x’ then we have
    ∴ (y) = (3x2 ) = 3 (x2) = 3 ×2 x = 6x
     
    Rules
    � Rule -4: [ f(x) ±g(x)] = f(x) ± g(x)
     
    Example: If y = 5x4 + 3x2 + 2x +7 then find y .
     
    Solution:
    Given
    y = 5x4 + 3x2 + 2x +7
    Differentiating w. r. t. ‘x’ then we have
    ∴ ( y ) = (5x4 + 3x2 + 2x +7)
    (y) = (5x4) + (3x2) + (2x) + (7)
    y = 5 (x4) + 3 (x2) + 2 (x) + (7)
    y = 20 x3 + 6 x + 2 .1 +0
    =20 x3 + 6x + 2 (Ans.)
    Rules
    �Problem:
    (i) If y = 2 x - 4 x3 + 5x – 10 then find y .
    (ii) If y = 7q3 – 4p2q2 +3q then find = ?
     
    Rules
    � Rule -5(Product rule): [ f(x) g(x)] = f(x) [g(x)] + g(x)
    [f(x)]
     
    Example – (i) If f(x) = (x2)(x3) then find f1(x) .
     
    Solution:
    Given f(x) = (x2)(x3)
    Differentiating w. r. t. ‘x’ then we have
    ∴ f(x) = (x2)(x3)
    = x2 x3 + x3 x2 = x2 × 3 x2 + x3 × 2x = 3x4 + 2x3
     
    Rules

    �Find the derivative of i)g(x)=(2x-7)10


    ii)g(x)=(3x2 -2x+5)3/2
    Rules
    (ii) If f(x) = (x2 +3)(2x + 5)3/2 then find f1(x).
     
    Solution:
    Given f(x) = (x2 +3)(2x + 5)3/2
    Differentiating w. r. t. ‘x’ then we have
    ∴ f(x) = (x2 +3)(2x + 5)3/2
    = (x2 +3) (2x + 5)3/2 + (2x + 5)3/2 (x2 +3)
    = (x2 +3) (2x + 5)3/2 (2x+5) + (2x + 5)3/2 [ (x2) + (3) ]
    = (x2 +3) (2x + 5) [ (2x) + (5)] + (2x + 5)3/2 (2x +0)
    = (x2 +3 )(2x + 5) (2 +0) + 2x (2x + 5)3/2
    = 3(x2 +3 )(2x + 5) + 2x (2x + 5)3/2
    Rules
    Rules

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