How do you find the derivative of #(2x^2 +x - 3)/x#?

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Answer 1 · 2016-01-18T16:35:58

#=(2x^2+3)/x^2#

The quotient rule states that for a function #y=f/g#, then #y'=(f'g-fg')/g^2#.

Here, we have #f=2x^2+x-3# and #f'=4x+1#. Additionally, #g=x# and #g'=1#.

Applying the quotient rule, we have the derivative equalling

#((4x+1)(x)-(2x^2+x-3)(1))/x^2#

#=(2x^2+3)/x^2#

Answer 2 · 2016-01-18T18:54:52

The derivative can be written: #2+3x^-2# or #2+3/x^2# or #(2x^2+3)/x^2#. All are equivalent.

This can clearly be differentiated using the quotient rule, but it is also possible to rewrite the expression before differentiating.

#(2x^2+x-3)/x = (2x^2)/x+x/x-3/x = 2x+1-3x^-1#

So the derivative is: #2+3x^-2#

This result can be written in other forms:

#2+3x^-2 = 2+3/x^2=(2x^2+3)/x^2#.