4 4.1 Differentiation The First Principle The above figure shows the graph of a general function y = f (x). M and N are points (a, 0) and (a + h, 0) respectively. P and Q are the points on the curve given by x = a and x = a + h. So M P = f (a) and N Q = f (a + h). It can be observed that RQ = N Q − N R = NQ − MP = f (a + h) − f (a) The gradient of PQ = = RQ PR f (a + h) − f (a) h Hence, the gradient at f (a + h) − f (a) . h Thus, the gradient of the function f (x) at any point on the curve is given by P = lim h→0 f (x + h) − f (x) . h f ′ (x) is called the derived function of f (x), and this method used in obtaining f ′ (x) is generally known as the first principle. f ′ (x) = Example 4.1. Use the principle method to find f ′ (x) given that (a) f (x) = x (b) f (x) = 2x2 x (c) f (x) = x+1 Solution. (a) Here, f (x) = x, f (x + h) = x + h. f (x + h) − f (x) (x + h) − x h = = =1 h→0 h h h f ′ (x) = lim (b) f (x) = 2x2 , ∴ f (x + h) = 2(x + h)2 = 2(x2 + 2hx + h2 ) = 2x2 + 4hx + 2h2 f (x + h) − f (x) f ′ (x) = lim h→0 h 2 2x + 4hx + 2h2 − 2x2 = lim h→0 h 4hx + 2h2 = lim h→0 h = lim 4x + 2h = 4x h→0 15 (h ΜΈ= 0). (c) Here, f (x) = x x+h , f (x + h) = x+1 x+h+1 f ′ (x) = = = = = = = = lim x+h x+h+1 − h h→0 lim x x+1 (x+h)(x+1)−x(x+h+1) (x+h+1)(x+1) h (x + h)(x + 1) − x(x + h + 1) lim h→0 h(x + h + 1)(x + 1) x2 + hx + x + h − x2 − hx − x lim h→0 h(x + h + 1)(x + 1) h lim h→0 h(x + h + 1)(x + 1) 1 lim h→0 (x + h + 1)(x + 1) 1 (x + 1)(x + 1) 1 (x + 1)2 h→0 Example 4.2. Find f ′ (x) by the first principle method in each of the following: 1 (a) f (x) = x 2 (b) f (x) = xn 1 (c) f (x) = x− 2 Solution. 1 (a) Here f (x) = x 2 = √ 2. f (x + h) = f ′ (x) = = = = = = √ x + h. f (x + h) − f (x) h→0 h √ √ x+h− x lim h→0 h √ √ √ √ x+h− x x+h+ x lim ×√ √ h→0 h x+h+ x x+h−x lim √ √ h→0 h( x + h + x) 1 lim √ √ h→0 x+h+ x 1 1 √ = x−1/2 2 2 x lim (b) f (x) = xn , f (x + h) = (x + h)n = xn + nxn−1 h + n(n − 1) n−2 2 x h + · · · + hn 2! 16 f (x + h) − f (x) h n x + nxn−1 h + n(n−2) xn−2 h2 + · · · + hn − xn 2 = lim h→0 h n(n−2) nxn−1 h + 2 xn−2 h2 + · · · + hn = lim h→0 h n(n − 2) n−2 n−1 n−1 + = lim nx x h + ··· + h h→0 2 f ′ (x) = lim h→0 = nxn−1 1 1 1 (c) Here, f (x) = x− 2 = √ and f (x + h) = √ . x x+h f ′ (x) = = = = = = = = f (x + h) − f (x) h→0 h √1 √1 − x x+h lim h→0 h √ √ x− x+h lim √ √ h→0 h( x x + h) √ √ √ √ x− x+h x+ x+h √ lim √ √ ×√ h→0 h( x x + h) x+ x+h x − (x + h) √ lim √ √ √ h→0 h( x x + h)( x + x + h) −1 √ lim √ √ √ h→0 ( x x + h)( x + x + h) −1 √ x×2 x −1 lim 3 2x 2 1 3 = − x− 2 2 4.1.1 Exercise 1. Find from the first principles the derived function of the following (a) y = x3 + 2x (b) f (x) = x2 + x (c) g(x) = 4x2 + 2 1 (d) h(x) = 3 2x (e) y = x4 − x2 1 (f) f (x) = 2x + 3 1 (g) y = 3 x 2. Find the derived function of the following: 17 (a) (x2 + 1)(x − 1) (b) (x + 21 )2 2 (c) 5 x 1 (d) √ 4x2 + 1 √ x (e) 3 x2 4.2 Techniques of Differentiation In the preceeding examples; we used the limit definition to find derivatives. The process of finding the derivative of a function is called differentiation. A function is differentiable at x if its derivative exists at x, and is differentiable on an interval (a, b) if it is differentiable at every point in the interval. In addition to f ′ (x), which is read as f prime of x. Other notations are dy ′ d used to denote the derivative of y = f (x). The most common are ,f, [f (x)] and Dx [y]. dx dx dy The notation is read as the derivative of y with respect to x or simply, dy − dx. Using the dx limit notation, we can write dy f (x + h) − f (x) = lim = f ′ (x). dx h→0 h In what follows, we introduce several rules that allow us to find derivatives without the direct use of the limit definition. We start with the simplest of all functions, the constant function f (x) = c. 4.3 Derivatives of a Constant Function The derivative of a constant function is 0. That is if f (x) = c, a real number, then d [c] = 0. dx For example, if y = 7, then 4.4 dy = 0. dx The Power Rule If n is any real number, then the function f (x) = xn is differentiable and d [f (x)] = nxn−1 . dx Example 4.3. Find dy if dx (a) y = x3 √ (b) y = x Solution. 18 (a) If y = x3 , then by the power rule, dy = 3x3−1 = 3x2 . dx 1 (b) We express the function in index form. That is, y = x 2 . Therefore, dy dx = = = 4.5 1 1 −1 x2 2 1 −1 x 2 2 1 1 √ 1 = 2 x 2x 2 The constant multiple rule If c is any constant and f (x) is a differentiable function, then d d [cf (x)] = c [f (x)]. dx dx Example 4.4. Differentiate the following (a) y = 6x4 (b) f (x) = 4 x2 Solution. dy (a) = 4(6)x4−1 = 24x3 dx (b) f (x) can be rewritten as f (x) = 4x−2 . Thus, f ′ (x) = −2(4)x−2−1 = −8x−3 8 = − 3 x 4.6 The Sum and Difference rule If f and g are both differentiable, then d d d [f (x) + g(x)] = [f (x)] + [g(x)] dx dx dx and d d d [f (x) − g(x)] = [f (x)] − [g(x)]. dx dx dx Example 4.5. Differentiate the following: (a) f (x) = −2x3 + 9x2 − 2 (b) g(t) = t4 + 3t3 − t2 − 13t 2 Solution. (a) f ′ (x) = −6x2 + 18x (b) g ′ (t) = 4t3 + 9t2 − 2t − 13 2 19 4.7 The Product Rule If f and g are both differentiable, then d d d [f (x)g(x)] = f (x) [g(x)] + g(x) [f (x)]. dx dx dx Example 4.6. Find the derivative of h(x) = (3x − 2x2 )(5 + 4x). Solution. d d [5 + 4x] + (5 + 4x) [3x − 2x2 ] dx dx = (3x − 2x2 )(4) + (5 + 4x)(3 − 4x) h′ (x) = (3x − 2x2 ) = (12x − 8x2 )(15 − 8x − 16x2 ) = −24x2 + 4x + 15 4.8 The Quotient Rule If f and g are differentiable, then d d g(x) dx [f (x)] − f (x) dx [g(x)] d f (x) = . 2 dx g(x) [g(x)] Example 4.7. If y = x2 + x − 2 dy , find 3 x +6 dx Solution. dy dx = = = = 4.8.1 d d (x3 + 6) dx [x2 + x − 2] − (x2 + x − 2) dx [x3 + 6] (x3 + 6)2 3 (x + 6)(2x + 1) − (x2 + x − 2)(3x2 ) (x3 + 6)2 (2x4 + x3 + 12x + 6) − (3x4 + 3x3 − 6x2 ) (x3 + 6)2 −x4 − 2x3 + 6x2 + 12x + 6 (x3 + 6)2 Exercise 1. Find y ′ in each of the following (a) y = 7x5 − 3x4 + x2 2 (b) y = 2t4 − 6t + t − 2 t √ (c) y = 6 3 x 2. Differentiate the following functions with respect to x: (a) f (x) = (x3 + 1)(x4 + 1) (b) g(x) = (x + x3 )(x2 − 1) 3. Find dy in the following: dx 20 x2 x+1 t3 + t (b) y = 4 t −2 x2 √ (c) y = 1+ x (a) y = 4.9 The Chain Rule If y = f (u) is a differentiable function of u, and u = g(x) is a differentiable function of x, then y = f (g(x)) is a differentiable function of x and dy dy du = × . dx du dx Example 4.8. Differentiate the following functions: √ (a) y = x2 + 1 (b) y = (3x2 − 4)4 Solution. (a) Using the chain rule, we let u = x2 + 1, then y = ∴ dy dx √ u and 1 dy du = √ , = 2x. du 2 u dx dy du × du dx 1 √ × 2x 2 u x x √ =√ 2 u x +1 = = = (b) Here again, using the chain rule, we let u = 3x2 − 4, then y = u4 and by the general power rule, dy du = 4u3 and = 6x. du dx Thus, dy dy du = × dx du dx = 4u3 × 6x = 4(3x2 − 4)3 × 6x = 24x(3x2 − 4)3 Example 4.9. If f (x) = 1 , find f ′ (x). (3x − 4)2 Solution. Note that you don’t need to use the quotient rule every time you see a quotient. Here we can rewrite the function as f (x) = (3x − 4)−2 . If we let u = 3x − 4, then f (u) = u−2 and by the general power rule, f ′ (u) = −2u−3 and u′ (x) = 3. 21 Thus, f ′ (x) = f ′ (u) × u′ (x) = −2u−3 × 3 = −6u−3 6 = − 3 u −6 = (3x − 4)3 4.10 Derivatives of Trigonometric Functions We now extend our technique of differentiation to include the trigonometric functions. We begin by stating a special theorem on trigonometric limits without proof, which shall be required later. Theorem 4.10. sin x 1 − cos x = 1 and lim =0 x→0 x x→0 x lim This result plays a crucial part in finding the derivatives of sin x and cos x. To obtain their derivatives, we must work in radians. 4.11 Derivative of sin x and cos x If f (x) = sin x, then f ′ (x) = = = = = = = = = = f (x + h) − f (x) h→0 h sin(x + h) − sin x lim h→0 h sin x cos h + cos x sin h − sin x lim h→0 h sin x cos h − sin x + cos x sin h lim h→0 h sin x(cos h − 1) + cos x sin h lim h→0 h cos h − 1 sin h lim sin x + cos x h→0 h h cos h − 1 sin h lim sin x + lim cos x h→0 h→0 h h cos h − 1 sin h + cos x lim sin x lim h→0 h→0 h h 1 − cos h sin h − sin x lim + cos x lim h→0 h→0 h h − sin x(0) + cos x(1) lim = cos x Thus, d [sin x] = cos x. dx 22 Similarly, if f (x) = cos x, then f ′ (x) = − sin x. That is, d [cos x] = − sin x. (Prove as an exercise). dx Example 4.11. Differentiate the following functions: (a) y = sin 3x cos x (b) f (x) = 2 (c) g(x) = sin(4x2 − 3) (d) y = x6 cos x + 2x5 Solution. (a) Here, we let u = 3x, then y = sin u by the chain rule. Thus, dy dx dy du × du dx = cos u × 3 = = 3 cos u = 3 cos 3x (b) Let f (x) = cos x 1 = cos x. Thus, 2 2 f ′ (x) = 1 1 × − sin x = − sin x. 2 2 (c) Let u = 4x2 − 3, then g(u) = sin u and by the chain rule, g ′ (x) = g ′ (u) × u′ (x) = − cos u × 8x = −8x × cos u = −8x cos(4x2 − 3) (d) dy dx d d d [cos x] + cos x [x6 ] + [2x5 ] dx dx dx = x6 × (− sin x) + cos x × 6x5 + 10x4 = x6 = −x6 sin x + 6x5 cos x + 10x4 = x4 (−x2 sin x + 6x cos x + 10) Example 4.12. Differentiate the following with respect to x. (a) tan x (b) cosec x (c) sin3 x 1 − cos x 2 (d) sin x Solution. 23