Question

How do you find the first and second derivatives of `y = (x^2 + 3) / x` using the quotient rule?

Answer 1

Apply the quotient rule twice to find

`d/dx(x^2+3)/x = (x^2-3)/x^2`

`d^2/dx^2(x^2+3)/x = 6/x^3`

Explanation:

The Quotient Rule states that

`d/dx f(x)/g(x) = (f'(x)g(x) - f(x)g'(x))/g^2(x)`

First Derivative:

`d/dx (x^2+3)/x = ((d/dx(x^2+3))x - (x^2+3)(d/dxx))/x^2`

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

`= (2x^2-x^2-3)/x^2`

`= (x^2-3)/x^2`

Second Derivative:

`d^2/dx^2 (x^2-3)/x = d/dx (x^2-3)/x^2`

`= ((d/dx(x^2-3))x^2-(x^2-3)(d/dxx^2))/(x^2)^2`

`= ((2x)x^2-(x^2-3)(2x))/x^4`

`= (2x^3-2x^3 + 6x)/x^4`

`= (6x)/x^4`

`= 6/x^3`

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As a side note, the problem could be made slightly easier by initially splitting the function into two terms:

`d/dx (x^2+3)/x = d/dx x + 3/x = 1 + ((d/dx3)x-3(d/dxx))/x^2 = 1-3/x^2`

`d/dx(1-3/x^2) = -((d/dx(3))x^2-3(d/dxx^2))/(x^2)^2 = -((-6x)/x^4) = 6/x^3`