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

##### 1 Answer

#### 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(g(x)))=1/(g(x))*g'(x)#

The given function is

#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)=((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)#