University:
Kentucky State University
Course:
MAT 120 | Precalculus
Academic year:

2024
Views:

32

Pages:

19
Author:

Madelyn Bradshaw

**Subject: Differentiation (Part 1)** **Date:** **No.** * **Definition:** For f(x), we define the first derivative of f at a by: ``` f'(a) = lim(x→a) (f(x)-f(a))/(x-a) ``` ``` f'(a) = lim(h→0) (f(a+h)-f(a))/h ``` * **Rules of Differentiation:** 1. d/dx(c) = 0 2. If f(x) = x^n, n∈R => f'(x) = nx^(n-1) 3. If f(x), g(x) exist => (f+g)'(x) = f'(x) + g'(x) * **Ex: Find f'(x)** 1. f(x) = x³ - 6x² + 5 f'(x) = 3x² - 12x 2. f(x) = 3x⁵ - x⁴ + x⁻¹ f'(x) = 15x⁴ - 4x³ - x⁻² 3. f(x) = 1/√x + 5√x f'(x) = -1/2x^(3/2) + 5/2x^(1/2) * **Remark:** If f'(x), g(x) exist, then: 1. (fg)'(x) = f(x)g'(x) + f'(x)g(x) 2. (f/g)'(x) = (g(x)f'(x) - f(x)g'(x)) / g²(x) 3. [(f(x))^n]' = n(f(x))^(n-1) * f'(x) 1. y = (x²+3x)².(2x+5) dy/dx = (x²+3x)².(4x+5) + (2x+5).(4(x²+3x).(2x+3)) 2. y = (x²+3)(x²-1)⁵ dy/dx = 5(x²+3)⁴.[(x+3)(2x) - (x²-1)(1)] / (x+3)² **Note:** * If f'(a) exists then f(x) is cont. at x=a * If f(x) is dis cont. at x=a then f'(a) not exist **eg)** If f(x) = x²cos(2/x), x≠0, is f'(0) exist? differentiable? f(0) = 0 lim (x→0) x²cos(2/x) = 0, since -1 ≤ cos(2/x) ≤ 1 -x² ≤ x²cos(2/x) ≤ x² 0 ← x²cos(2/x) ← 0 lim(x→0) f(x) = 0 = f(0) ∴ f is cont. f'(0) d.n.e **3. d/dx [π^(3x+1)] = π^(3x+1) . 3 . lnπ** **4. If y = x^sin(x), find dy/dx** lny = sin(x) ln(x) (1/y) dy/dx = sin(x)/x + ln(x)cos(x) dy/dx = y [sin(x)/x + ln(x)cos(x)] dy/dx = x^sin(x) [sin(x)/x + ln(x)cos(x)] **Table:** x f(x) f'(x) g(x) g'(x) 1 3 0 2 5 2 5 4 3 1 3 1 6 1 0 **5. If h(x) = f(g(x)), then h'(1) = ?** h'(1) = f'(g(1)) * g'(1) = f'(2) * 7 = 4 * 7 = 28 **6. If h(x) = √(3 + f(x)), then h'(2) = ?** h'(2) = f'(2) / (2√(3+f(2))) = 4 / (2√(3+1)) = 1 **7. If h(x) = g(x)cos(πx/4), then h'(2) = ?** h'(2) = g(2)(-sin(π/2)(π/4)) + cos(π/2)(g'(2)) = 3(-sin(π/2)(π/4)) + 0(1) = 3(-1)(π/4) = -3π/4 **Differentiation Part (4)** * **Differentiation of Inverse Function:** If y = f⁻¹(x) is one-to-one, then f'(x) exists & ``` d/dx[f⁻¹(x)] = 1 / f'(f⁻¹(x)) ``` If f'(a) = 0, then (f⁻¹)'(a) does not exist. **eg) If f(x) = x³ + 3x² + 2, find (f⁻¹)'(6)** 1. f'(x) = 3x² + 6x 2. f'(1) = 3 + 6 = 9 3. (f⁻¹)'(6) = 1 / f'(1) = 1/9 **eg) If f(x) = 3e²ˣ + x³, find (f⁻¹)'(11)** 1. f'(x) = 6e²ˣ + 3x² 2. f'(2) = 6e⁴ + 12 = 11 3. (f⁻¹)'(11) = 1 / f'(2) = 1 / (6e⁴ + 12) **Diff of Inverse trig. function:** 1. dx/d[sin⁻¹(g(x))] = g'(x) / √(1-g²(x)) 2. dx/d[cos⁻¹(g(x))] = -g'(x) / √(1-g²(x)) 3. dx/d[tan⁻¹(g(x))] = g'(x) / (1+g²(x)) 4. dx/d[cot⁻¹(g(x))] = -g'(x) / (1+g²(x)) 5. dx/d[sec⁻¹(g(x))] = g'(x) / |g(x)|√(g²(x)-1) 6. dx/d[cosec⁻¹(g(x))] = -g'(x) / |g(x)|√(g²(x)-1) **eg) Find dy/dx** 1. y = tan⁻¹x + sin⁻¹x dy/dx = 1/(1+x²) + 1/√(1-x²) 2. y = sin⁻¹(2x) + 3cos⁻¹(x²) dy/dx = 2/√(1-4x²) + 3(-2x)/√(1-x⁴) = 2/√(1-4x²) - 6x/√(1-x⁴) 3. y = sec⁻¹(lnx) dy/dx = (1/|lnx|)√((lnx)²-1) * (1/x)