Question

Find the derivative of
y= (x^3+4x-2)/(x^3)

Answer 1

Use u/v rule (quotient rule)

Answer 2

\[y=\frac{ x ^{3}+4x-2 }{x ^{3} }=1+4x ^{-2}-2x ^{-3}\]
\[\frac{ dy}{dx }=0+4*-2*x ^{-3}-2*-3x ^{-4}=\frac{ -8 }{x ^{3} }+\frac{ 6 }{ x ^{4} }\]

Answer 3

Use the quotient rule, which is:\[\Large \frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\]\[\Large y=\frac{x^3+4x-2}{x^3}\]\[\Large \frac{dy}{dx}=\frac{(3x^2+4)(x^3)-(x^3+4x-2)(3x^2)}{x^6}\]\[\Large \frac{dy}{dx}=\frac{(3x^5+4x^3)-(3x^5+12x^3-6x^2)}{x^6}\]\[\Large \frac{dy}{dx}=\frac{6x^2-8x^3}{x^6}=\frac{6}{x^4}-\frac{8}{x^3}\]