This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Problem 1a: f(x) = x^2 x Step 1: Product rule (uv)' = u'v + uv', where u = x^2, v = x. u' = 2x, v' = x. f'(x) = 2x x + x^2 x. f'(x) = 2x x + x^2 x f'(x) = 2x x + x^2 x Problem 1b: y = (x)/(x^2) + 1 Step 1: Quotient rule ( (u)/(v) )' = (u'v - uv')/(v^2), where u = x, v = x^2 + 1. u' = 1, v' = 2x. y' = (1)(x^2 + 1) - x(2x)(x^2 + 1)^2. Step 2: Simplify numerator: x^2 + 1 - 2x^2 = 1 - x^2. y' = 1 - x^2(x^2 + 1)^2. y' = 1 - x^2(x^2 + 1)^2 y' = 1 - x^2(x^2 + 1)^2 Problem 1c: f(x) = e^x x Step 1: Product rule (uv)' = u'v + uv', where u = e^x, v = x. u' = e^x, v' = - x. f'(x) = e^x x + e^x (- x) = e^x ( x - x). f'(x) = e^x ( x - x) f'(x) = e^x ( x - x) Problem 1d: y = sqrt(x^2) + 1 Step 1: Rewrite y = (x^2 + 1)^1/2. Chain rule y' = (1)/(2) (x^2 + 1)^-1/2 · 2x = (x)/(sqrt(x^2) + 1). y' = (x)/(sqrt(x^2) + 1) y' = (x)/(sqrt(x^2) + 1) Problem 2a: y = ^3 x Step 1: Chain rule, let u = x, y = u^3. (dy)/(dx) = 3u^2 (du)/(dx) = 3 ^2 x x. y' = 3 ^2 x x y' = 3 ^2 x x Problem 2b: y = ( x)/( x) Step 1: Rewrite y = ( x / x)/(1 / x) = x. y' = x. y' = x y' = x Problem 3: Tangent line to y = x^2 at x = 1. Step 1: y' = 2x, slope at x=1: m = 2(1) = 2. Point: (1, 1). Equation: y - 1 = 2(x - 1). y = 2x - 1. y = 2x - 1 Problem 4: Implicit x^2 + y^2 = 25, find (dy)/(dx). Step 1: Differentiate: 2x + 2y y' = 0. 2y y' = -2x. y' = -(x)/(y). (dy)/(dx) = -(x)/(y) (dy)/(dx) = -(x)/(y) Problem 5: Ladder 10 ft long against wall. Bottom slides away at 1 ft/s, top speed when bottom 6 ft from wall. Step 1: Let x distance from wall (bottom), y height on wall (top). x^2 + y^2 = 10^2 = 100. Differentiate: 2x (dx)/(dt) + 2y (dy)/(dt) = 0. (dy)/(dt) = -(x)/(y) (dx)/(dt). When x=6 ft, y = sqrt(100 - 36) = 8 ft. (dx)/(dt) = 1 ft/s. (dy)/(dt) = -(6)/(8) (1) = -0.75 ft/s. -0.75 ft/s