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

1 Answer
Feb 17, 2016

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)