University:
California State University, Northridge
Course:
MATH 351 | Differential Equations
Academic year:

2024
Views:

314

Pages:

3
Author:

Javon Neal

Question 1 (a) Find the differential dy. y = √(24 + x²) (b) Evaluate dy for the given values of x and dx. x = 1, dx = –0.2 Answer Find the derivative using chain rule. Then multiply both sides by dx. (a) y = √(24 + x²) y = (24 + x²)¹/² dy/dx = 1/2 (24 + x²)⁻¹/² (2x) = x / √(24 + x²) dy = x / √(24 + x²) dx Substitute x = 1 and dx = –0.2 into the differential from part (a). Then evaluate. (b) dy = x / √(24 + x²) dx = 1 / √(24 + 1²) (–0.2) = 1 / √25 (–0.2) = 1/5 (–0.2) = 0.2 * (–0.2)* = –0.04 Results: (a) dy = x / √(24 + x²) dx (b) dy = –0.04 Question 2 Find the derivatives of the following function f(x): f(x) = (√(x) – 3x) (x + 1/x) Answer Step 1 Given: f(x) = (√x – 3x) (x + 1/x) Step 2 Differentiating with respect to x: f'(x) = (√x – 3x) d/dx (x + 1/x) + (x + 1/x) d/dx (√x – 3x) = (√x – 3x) (1 – 1/x²) + (x + 1/x) (1 / (2√x) – 3) Simplify step by step: f'(x) = √x – 3x + 3x/x² – 3x² + x / (2√x) – 3/x + x²/√x = Final result: f'(x) = 3 / √x – 6x Question 3 Find the derivative of the problem. h(x) = sin(2x cos(2x)) Answer Product Rule of Differentiation: If f(x) and g(x) are two differentiable functions, then: (fg)' = f'(x)g(x) + f(x)g'(x) The given function is h(x) = sin(2x cos(2x)) Differentiate using product rule: h'(x) = d/dx [sin(2x cos(2x))] = sin(2x) d/dx (cos(2x)) + cos(2x) d/dx (sin(2x)) = –2sin(2x) + 2cos(2x) cos(2x) = –2 + 4cos²(2x) Question 4 Use implicit differentiation to find ∂z / ∂x and ∂z / ∂y. e⁴z = xyz Answer Given: e⁴z = xyz Step 1: Partial differentiating with respect to x: d/dx (e⁴z) = d/dx (xyz) e⁴z dz/dx = yz dz/dx = yz / e⁴z – xy Step 2: Partial differentiating with respect to y: d/dy (e⁴z) = d/dy (xyz) e⁴z dz/dy = xz dz/dy = xz / e⁴z – xy Question 5 Find the derivative of the function: y = ln(sec x + tan x) Answer y = ln(sec x + tan x) Differentiate with respect to x: dy/dx = 1 / (sec x + tan x) * d/dx (sec x + tan x) = 1 / (sec x + tan x) (sec x tan x + sec² x) = (sec x tan x + sec² x) / (sec x + tan x) Result: dy/dx = sec x

Derivatives #20 - Questions and Answers

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