How to solve Differentiate: Quotient rule?

Question

1. y=(3x)/(x+5)

2. y= (x)/(2x^2 + 1)

3. y= (x^3 + x)/(x^2 - x - 1)

4. y= (3x + 4)^5/(2x - 5)^3

5. y= sqrt(x-1)/(2x-3)

Answer 1

1. quotient rule

2. quotient rule

3. quotient rule

4. quotient rule with chain rule

think quotient rule first

y = f/g

y' = (f'g - fg')/g^2

y' = {[5(3x+4)^4](3)(2x-5)^3 - [(3x+4)^5][3(2x-5)^2](2)}/(2x-5)^6

5. same as 4 but with sqrt(x-1) = (x-1)^1/2

Answer 2

Quotient rule: f(x) = g(x)/h(x)

d/dx f(x) = f'(x) = [g'(x)h(x) - g(h)h'(x)] / [h(x)]²

1. g(x) = 3x ; g'(x) = 3 ; h(x) = (x+5) ; h'(x) = 1

f'(x) = [(3)(x+5) - (3x)(1)] / (x+5)² = 15 / (x+5)²

Answer: 15 / (x+5)²

2. g(x) = x ; g'(x) = 1 ; h(x) = (2x² + 1) ; h'(x) = 4x

f'(x) = [(1)(2x² + 1) - x(4x)] / (2x² + 1)² = (1 - 2x²) / (2x² + 1)²

Answer: (1 - 2x²) / (2x² + 1)²

3. g(x) = (x³ + x) ; g'(x) = 2x² + 1 ; h(x) = (x² -x -1) ; h'(x) = 2x  - 1

f'(x) = [(2x² + 1)(x² -x -1) - (x³ + x)(2x  - 1)] / (x² -x -1)²

f'(x) = (x^4 - 2x³ - 4x² - 1)/(x² -x -1)²

Answer: (x^4 - 2x³ - 4x² - 1)/(x² -x -1)²

4. g(x) = (3x + 4)^5 ; g'(x) = 15(3x + 4)^4 ; h(x) = (2x - 5)^3 ; h'(x) = 6(2x-5)²

f'(x) = [15(3x + 4)^4((2x - 5)^3 - (3x + 4)^5(6)((2x-5)²] / (2x - 5)^6

f'(x) = [15(3x + 4)^4((2x - 5) - 6((3x + 4)^5] / (2x - 5)^4

f'(x) = 3(3x + 4)^4 [ 5(2x-5) - 2(3x+4)] / (2x - 5)^4

f'(x) = 3(3x + 4)^4 (4x - 33) / (2x - 5)^4

Answer: 3(3x + 4)^4 (4x - 33) / (2x - 5)^4

5. g(x) = √(x-1) ; g'(x) = 1/2√(x-1) ; h(x) = (2x - 3) ; h'(x) = 2

f'(x) = [(2x - 3)1/2√(x-1) - 2√(x-1)] / (2x - 3)²

f'(x) = (1 - 2x) / [2(2x - 3)²√(x-1)]

Answer: (1 - 2x) / [2(2x - 3)²√(x-1)]