Question

Question  2

Find the derivative of the function Y= x³ + x² /x

a. x2/x + x/x

b. x2+ x /x

c. 2x + 1

d. 3x⁴ + 2x³

Answer 1

$y = \dfrac{x³ + x²}{x}= \\ u = x³ + x²\\ du/dx = 3x² + 2x\\ v = x \\dv/dx = 1 \\ \dfrac{dy}{dy} = \dfrac{Vdu/dx-Udv/dx}{V²}$

$\dfrac{dy}{dx} =\dfrac{x(3x²+2x)-(x³+x²)(1)}{x²}$

$= \dfrac{3x³+2x²-x³-x²}{x²} =\dfrac{2x³+x²}{x²}$

$=2x+1$

$a.\ \ y = x +1\\ \quad \dfrac{dy}{dx} = 1(x⁰) = 1$

$b.\ \ y = 2x +1\\ \quad \dfrac{dy}{dx} = 2(x⁰) +0 = 2$

$c.\ \ y = 2x +1\\ \quad \dfrac{dy}{dx} = 2(x⁰)+ 0 = 2$

$d.\ \ y = 3x⁴ +2x³\\ \quad \dfrac{dy}{dx} = 3(4x³) + 6(x²) = 12x³ + 6x²$