Question

Find derivative of x+1/x whole cube

Answer 1

Answer:

$\frac{d}{dx}$(x³ + x⁻³ + 3x + 3x⁻¹) = 3x² - 3x⁻⁴ + 3 - 3x⁻²

Step-by-step explanation:

(x + $\frac{1}{x}$)³ = x³ + 1/x³ + 3(x + 1/x)

           = x³ + x⁻³ + 3x + 3x⁻¹

$\frac{d}{dx}$(x³ + x⁻³ + 3x + 3x⁻¹)

= $\frac{d}{dx}$($x^{3}$) + $\frac{d}{dx}(x^{-3}) + \frac{d}{dx}(3x) + \frac{d}{dx}(3x^{-1})$

= 3$x^{3-1}$ + (-3)($x^{-3-1}$) + 3(1) + 3(-1)($x^{-1-1}$)

= 3x² - 3x⁻⁴ + 3 - 3x⁻²