Answer:
[tex] \displaystyle \frac{dy}{dx} = \frac{2x + 2}{x^3} [/tex]
Step-by-step explanation:
we would like to figure out the derivative of the following:
[tex] \displaystyle \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} } [/tex]
to do so, let,
[tex] \displaystyle y = \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} } [/tex]
By simplifying we acquire:
[tex] \displaystyle y = 3 - \frac{2}{x} - \frac{1}{ {x}^{2} } [/tex]
use law of exponent which yields:
[tex] \displaystyle y = 3 - 2 {x}^{ - 1} - { {x}^{ - 2} } [/tex]
take derivative in both sides:
[tex] \displaystyle \frac{dy}{dx} = \frac{d}{dx} (3 - 2 {x}^{ - 1} - { {x}^{ - 2} } )[/tex]
use sum derivation rule which yields:
[tex] \rm\displaystyle \frac{dy}{dx} = \frac{d}{dx} 3 - \frac{d}{dx} 2 {x}^{ - 1} - \frac{d}{dx} {x}^{ - 2} [/tex]
By constant derivation we acquire:
[tex] \rm\displaystyle \frac{dy}{dx} = 0 - \frac{d}{dx} 2 {x}^{ - 1} - \frac{d}{dx} {x}^{ - 2} [/tex]
use exponent rule of derivation which yields:
[tex] \rm\displaystyle \frac{dy}{dx} = 0 - ( - 2 {x}^{ - 1 -1} ) - ( - 2 {x}^{ - 2 - 1} )[/tex]
simplify exponent:
[tex] \rm\displaystyle \frac{dy}{dx} = 0 - ( - 2 {x}^{ -2} ) - ( - 2 {x}^{ - 3} )[/tex]
two negatives make positive so,
[tex] \displaystyle \frac{dy}{dx} = 2 {x}^{ -2} + 2 {x}^{ - 3} [/tex]
by law of exponent we acquire:
[tex] \displaystyle \frac{dy}{dx} = \frac{2 }{x^2}+ \frac{2}{x^3} [/tex]
simplify addition:
[tex] \displaystyle \frac{dy}{dx} = \frac{2x + 2}{x^3} [/tex]
and we are done!
