Calculus Derivative Solutions and Function Analysis

Derivatives of Algebraic and Transcendental Functions

• f(x) = (x³ - 3x)³: f'(x) = 3(x³ - 3x)² · (3x² - 3) = 9x⁸ - 63x⁶ + 135x⁴ - 81x²
• f(x) = 2e^x · (3x - 1): f'(x) = 2e^x · (3x - 1) + 2e^x · 3 = 6xe^x - 2e^x + 6e^x = 6xe^x + 4e^x
• f(x) = ln(x² + 6x): f'(x) = (1 / (x² + 6x)) · (2x + 6)
• f(x) = 1 / ln x: f'(x) = -1(ln x)^-2 · (1 / x) = -1 / (x ln² x)
• f(x) = 2 cos x · sin x: f'(x) = 2(-sin x) · sin x + 2 cos x · cos x = 2(cos² x - sin² x)
• f(x) = 2^{x2 - 3}: f'(x) = 2^{x2 - 3} · 2x · ln 2
• f(x) = cos³(2x): f'(x) = 3 · (cos 2x)² · (-sin 2x) · 2 = -6 cos²(2x) · sin(2x)
• f(x) = (2x³ - 3x) / (x² - 1): f'(x) = [(6x² - 3)(x² - 1) - (2x³ - 3x)(2x)] / (x² - 1)²
• f(x) = (x² - 1) / (x + 1): f'(x) = [(2x)(x + 1) - (x² - 1) · 1] / (x + 1)²
• f(x) = e^x2 / (1 - e^x): f'(x) = [e^x2 · 2x · (1 - e^x) + e^x2 · e^x] / (1 - e^x)² = (2xe^x2 - 2xe^x2+x + e^x2+x) / (1 - e^x)²
• f(x) = √(x² - 3x ln x): f'(x) = ln(x² - 3x)^1/2
• f(x) = ln √(x² - 3x): f'(x) = [1 / √(x² - 3x)] · (2x - 3) / [2√(x² - 3x)] = (2x - 3) / [2(x² - 3x)]
• f(x) = (2x² - e^x) / x: f'(x) = (4x - e^x · x - (2x² - e^x)) / x²
• f(x) = e^x2 · ln x²: f'(x) = e^x2 · 2x · ln x² + e^x2 · (2x / x²) = e^x2 · 2x (ln x² + (1 / x²))
• f(x) = e^x2 · (x² - 3x): f'(x) = e^x2 · 2x(x² - 3x) + e^x2(2x - 3) = e^x2(2x³ - 6x² + 2x - 3)
• f(x) = (x³ - 2x)³: f'(x) = 3(x³ - 2x)² · (3x² - 2)
• f(x) = 1 / (2√x): f(x) = (1/2)x^-1/3, f'(x) = (-1/6)x^-4/3 = -1 / (6√x⁴)
• f(x) = sin²(2x): f'(x) = 2 sin(2x) · cos(2x) · 2 = 2(2 sin 2x cos 2x) = 2 sin 4x
• f(x) = ln √(3x - 2): f'(x) = [1 / √(3x - 2)] · [1 / (2√(3x - 2))] · 3 = 3 / [2(3x - 2)]
• f(x) = (x² - 1) / (x + 1): f'(x) = [2x(x + 1) - (x² - 1) · 1] / (x + 1)² = [2x² + 2x - x² + 1] / (x + 1)² = 1
• f(x) = xe^x · ln x: f'(x) = (e^x + xe^x) ln x + xe^x · (1/x) = e^x(1 + x ln x + 1) = e^x(2 + x ln x)
• f(x) = (e^x - x) / e^x: f'(x) = [(e^x - 1)e^x - (e^x - x)e^x] / (e^x)² = (x - 1) / e^x

Tangent Line Calculation

Calculate the tangent line to: y = 3x² - 6x + 1, given the slope m = 12.

y' = 6x - 6. Setting 6x - 6 = 12, we find x = 3.
When x = 3, y = 3(3)² - 6(3) + 1 = 10. The point is (3, 10).
Using the point-slope form: y - 10 = 12(x - 3), which simplifies to y = 12x - 26.

Critical Points and Extrema Analysis

Analysis of f(x) = x³ - 3x² + 6

f' = 3x² - 6x = 0 → 3x(x - 2) = 0, so x = 2 and x = 0.
f(2) = (2)³ - 3(2)² + 6 = 2. Point: (2, 2).
f(0) = 6. Point: (0, 6).
f'' = 6x - 6. f''(2) = 6 > 0 (Relative Minimum); f''(0) = -6 < 0 (Relative Maximum).
Setting f'' = 6x - 6 = 0, we find x = 1, y = 4. Inflection point: (1, 4).

Analysis of f(x) = x²e^x

f'(x) = 2xe^x + x²e^x = e^x(x² + 2x) = 0.
x² + 2x = 0 → x(x + 2) = 0, so x = 0 and x = -2.
For x = 0, y = 0. Point: (0, 0) Minimum.
For x = -2, y = 4e^-2. Point: (-2, 4/e²) Maximum.
f''(x) = e^x(x² + 2x) + e^x(2x + 2) = e^x(x² + 4x + 4).
f''(0) = e⁰(4) > 0 (Minimum); f''(-2) = e^-2(0). (Note: Re-evaluating the original text's second derivative logic for extrema).