What is the derivative of #(x^2-2)/(x)#?

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Answer 1 · 2016-01-28T13:19:08

#frac{d}{dx}(frac{x^2-2}{x}) = 1 + 2/x^2#

#frac{d}{dx}(frac{x^2-2}{x}) = frac{d}{dx}(x-2/x)#

# = 1 + 2/x^2#

Answer 2 · 2016-01-28T13:31:31

#D = (x^2 -2) /x^2#

It states if we have two functions #u and v# then their derivative is given by

#D u/v = (v (d/(du)) u - u (d/(dv)) v)/ v^2 #

Here #u and v# are #(x^2 -2) and x# respectively.

Then,

#D (x^2-2)/x = [x d/dx (x^2-2) - (x^2-2) d/dx x] / x^2 #

# = (x * 2x - (x^2-2) * 1)/x^2 #

The constant is directly removed.

#= [2x^2 - (x^2 - 2)]/ x^2#

#D = (x^2 -2) /x^2#

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