Question

What is the second derivative of #f(x) = ln (x-x^2) #?

Answer 1

`f''(x)=(-2x^2+2x-1)/(x-x^2)`

Explanation:

To differentiate the natural logarithm function, use the chain rule. The chain rule states that

`d/dx(f(g(x)))=f'(g(x))*g'(x)`

Knowing that `d/dx(ln(x))=1/x`, we can apply this specifically to natural logarithm functions:

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

The given function is `f(x)=ln(x-x^2)`, so

`f'(x)=1/(x-x^2)*d/dx(x-x^2)`

`f'(x)=1/(x-x^2)*(1-2x)`

`f'(x)=(1-2x)/(x-x^2)`

To find the second derivative, use the quotient rule, which states that

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

Applying this to `f'(x)=(1-2x)/(x-x^2)`, we see that

`f''(x)=((x-x^2)d/dx(1-2x)-(1-2x)d/dx(x-x^2))/(x-x^2)^2`

Simplify through the power rule.

`f''(x)=((x-x^2)(-2)-(1-2x)(1-2x))/(x-x^2)^2`

`f''(x)=(-2x+2x^2-(1-4x+4x^2))/(x-x^2)^2`

`f''(x)=(-2x^2+2x-1)/(x-x^2)`