Question

What are the first and second derivatives of `g(x) =(lnx)^2-ln(x^2)`?

Answer 1

`d/dxg(x) = (2(ln(x)-1))/x`

`d^2/dx^2g(x) = (4-2ln(x))/x^2`

Explanation:

We will use

• The chain rule:
  `d/dxf(g(x)) = f'(g(x)g'(x)`

• The quotient rule:
  `d/dxf(x)/g(x) = (f'(x)g(x) - f(x)g'(x))/g^2(x)`

• `d/dx (f(x) +- g(x)) = f'(x) +- g'(x)`

• `d/dxln(x) = 1/x`

• `d/dx x^n = nx^(n-1)`

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

`=2ln(x)(d/dxln(x)) - 1/(x^2)(d/dxx^2)` (by the chain rule)

`= 2ln(x)(1/x) - 1/x^2(2x)`

`= (2ln(x))/x - 2/x`

`= (2(ln(x) - 1))/x`

To find the second derivative, we take the derivative of `g'(x)` using the quotient rule.

`g''(x) = d/dx(2(ln(x)-1))/x`

`= ((d/dx2(ln(x)-1))x - 2(ln(x)-1)(d/dxx))/x^2`

`=((2(1/x))x - 2(ln(x)-1)(1))/x^2`

`= (2 - 2ln(x) + 2)/x^2`

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