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Derivative of

\frac{ {3x}^(2) - 2x - 1 }{ {x}^(2) }
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Answer:


\displaystyle (dy)/(dx) = (2x + 2)/(x^3)

Explanation:

we would like to figure out the derivative of the following:


\displaystyle \frac{ { 3x }^(2) - 2x - 1 }{ {x}^(2) }

to do so, let,


\displaystyle y = \frac{ { 3x }^(2) - 2x - 1 }{ {x}^(2) }

By simplifying we acquire:


\displaystyle y = 3 - (2)/(x) - \frac{1}{ {x}^(2) }

use law of exponent which yields:


\displaystyle y = 3 - 2 {x}^( - 1) - { {x}^( - 2) }

take derivative in both sides:


\displaystyle (dy)/(dx) = (d)/(dx) (3 - 2 {x}^( - 1) - { {x}^( - 2) } )

use sum derivation rule which yields:


\rm\displaystyle (dy)/(dx) = (d)/(dx) 3 - (d)/(dx) 2 {x}^( - 1) - (d)/(dx) {x}^( - 2)

By constant derivation we acquire:


\rm\displaystyle (dy)/(dx) = 0 - (d)/(dx) 2 {x}^( - 1) - (d)/(dx) {x}^( - 2)

use exponent rule of derivation which yields:


\rm\displaystyle (dy)/(dx) = 0 - ( - 2 {x}^( - 1 -1) ) - ( - 2 {x}^( - 2 - 1) )

simplify exponent:


\rm\displaystyle (dy)/(dx) = 0 - ( - 2 {x}^( -2) ) - ( - 2 {x}^( - 3) )

two negatives make positive so,


\displaystyle (dy)/(dx) = 2 {x}^( -2) + 2 {x}^( - 3)

further simplification if needed:

by law of exponent we acquire:


\displaystyle (dy)/(dx) = (2 )/(x^2)+ (2)/(x^3)

simplify addition:


\displaystyle (dy)/(dx) = (2x + 2)/(x^3)

and we are done!

User Avimoondra
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