unit 1 differential calculus - IGNOU

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Mathematics-II 48 1.6.7 Derivatives of a Function Represented Parametrically Sometimes we represent a function say y = f (x), a ≤ x ≤ b by its parametric equations, x = φ( t) ⎤ y = ψ t ⎥, ( ) ⎦ α ≤ t ≤ β Let us assume φ (t), ψ (t) are differentiable and x = φ (t) has an inverse t = h (x). Then we can consider the equation of the function y = f (x) as a composite function. y = ψ( t), t = h( x) Using the differentiation rule for a composite function (chain-rule), we get dy dx dy = . dt dt dx dψ = dt dh dx dψ = dt dφ dt Since t = h (x) is inverse of x = φ (t), we arrive at Example 1.29 dy dx dψ = dt . dφ dt The parametric equation of a semi-circular arc of radius ‘a’ with its centre at (0, 0) is Find Solution dy dt x = a cos t⎤ y = a t ⎥, sin ⎦ π at t = . 4 dy dx = dy dt dx dt 0 ≤ t ≤ π a cos t = = − cot t a( − sin t) π dy π At t = , = − cot = − 1 4 dx 4 1.6.8 Logarithmic Differentiation In some problems, it is easier to find dx dy by first taking logarithmic and then differentiating. Such process is called logarithmic differentiation. This is usually done in two kinds of problems. First when the function is a product of many simpler functions. In this case logarithm converts the product into a sum and
facilitates differentiation. Secondly, when the variable x occurs in the exponent. In this case logarithm brings it to a simpler form. Example 1.30 Solution Solution Differentiate sin x sin 2x sin 3x. Let y = sin x sin 2x sin 3x Then log y = log sin x + log sin 2x + log sin 3x. Differentiating both sides, we have 1 y dy dx = 1 1 1 . cos x + . cos 2x . 2 + . cos3x . 3 sin x sin 2x sin 3x dy ⎡cos x 2cos 2x cos3x ⎤ ∴ = y ⎢ + + 3 ⎥ dx ⎣ sin x sin 2x sin 3x ⎦ Example 1.31 Differentiate (sin x) x Let y = (sin x) x Then log y = x log sin x = sin x sin 2x sin 3x [cot x + 2 cot 2x + 3 cot 3x] Differentiating both the sides, we have 1 y dy dx = log sin x + x . 1 . cos x sin x dy i.e. = y [log sin x + x cot x] dx = (sin x) x [log sin x + x cot x] 1.6.9 Differentiation by Substitution Sometimes it is easier to differentiate by making substitution. Usually these examples involve inverse trigonometric functions. Example 1.32 Solution − Differentiate ⎛ ⎞ ⎜ + x − x⎟ ⎝ ⎠ 2 1 tan 1 Put x = tan θ Then 1 + x 2 = sec 2 θ Differential Calculus 49
Page 1 and 2: UNIT 1 DIFFERENTIAL CALCULUS Struct
Page 3 and 4: Definition 2 Let S be a non-empty s
Page 5 and 6: 1 1 (ii) = , if x ≠ 0 x | x| (iii
Page 7 and 8: The role that the graph of a functi
Page 9 and 10: The Greatest Integer Function Figur
Page 11 and 12: Consider the function f : N → E d
Page 13 and 14: (iii) f : R → R defined by f (x)
Page 15 and 16: Remark (iv) Define a function q on
Page 17 and 18: ⎛ x 3 ⎞ ⎛ x 3 ⎞ fog (x) = f
Page 19 and 20: 0 Figure 1.13 (ii) The function g d
Page 21 and 22: Solution Look at Tables 1.1 and 1.2
Page 23 and 24: if, and only if, for each ε > 0 th
Page 25 and 26: (vi) lim x →a [ lim f ( x)] ⎡ f
Page 27 and 28: SAQ 4 lim = x →0 Evaluate the lim
Page 29 and 30: Remark If the interval is closed, t
Page 31 and 32: Example 1.20 Solution Draw the grap
Page 33 and 34: 1.6.1 Derivative of a Function We n
Page 35 and 36: Remark f ( 0 + h) − f ( 0) h −
Page 37 and 38: (ii) (iii) d dg ( x) df ( x) ( f (
Page 39 and 40: 1.6.3 Derivative of a Composite Fun
Page 41 and 42: (iii) (x 2 + x + 6) sin x (iv) (x +
Page 43: dy x Therefore, = y log e a = a log
Page 47 and 48: (c) Differentiate the following (i)
Page 49 and 50: SAQ 5 SAQ 6 SAQ 7 SAQ 8 (viii) Limi
Page 51: MATHEMATICS-II The knowledge and co

unit 1 differential calculus - IGNOU

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