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

Let f(x) = u/v such that:

u= x^3 + 1 ==> u' = 3x^2

v= 1-2x^2 ==> v'= -4x

==> f'(x) = (u'v-uv')/v^2

= (3x^2)(1-2x^2)- (x^3 +1)(-4x)]/(1-2x^2)^2

= (3x^ 2- 6x^4 + 4x^4 + 4x)/(1-2x^2)^2

= (-2x^4 + 3x^2 +4x)/(1-2x^2)^2

** = -x(2x^3 -3x -4)/(1-2x^2)^2**

To find the derivative of (x^3-1)/(1-2x^2).

We use {u(x)/v(x)}' = {u'(x)v(x) - u(x)v'(x)}/(v(x))^2 to find the derivative of the given function.

u(x) = x^3-1. So u'(x) = (x^3-1)' = 3x^2

v(x) = 1-2x^2. So v'(x) = (1-2x^2)' = -2*2x = -4x.

Therefore {(x^3-1)/(1-2x^2)}' = {(x^3-1)'(2x^2-1) - (x^3-1)(1-2x^2)'}/(1-2x^2)^2

= {(3x^2)(2x^2-1)-(x^3-1)(-4x)}/(1-2x^2)^2

{6x^4-3x^2-4x^4+4x)/(1-2x^2)^2

=(2x^4-3x^2+4x)/(1-2x^2)^2

=x(2x^3-3x+4)/(1-2x^2)^2

{(x^3-1)/(1-2x^2)}' = x(2x^3-3x+4)/(1-2x^2)

We find the derivative of the expression using the quotient rule.

Now for finding the derivative of f(x)/g(x) the quotient rule is [f’(x) g(x) – f(x) g’(x)]/ [g(x)] ^2

Here f(x) = x^3+1. So we have f’(x) 3x.

g(x) = 1- 2x^2, so g’(x) = -4x

Therefore the derivative of (x^3 + 1) / (1 - 2x^2 ) is

[(1- x^2) * 3x – (-4x) * (x^3 +1)] / (1- 2x^2) ^2

=> [(1- x^2) * 3x + 4x * (x^3 +1)] / (1- 2x^2) ^2

=> [3x - 3x^3 + 4x^4 + 4x] / (1- 2x^2) ^2

=> [ 4x^4 - 3x^3 + 7x] / (1- 2x^2) ^2

=> x*(4x^3 - 3x^2 +7) / (1- 2x^2)^2

**Therefore the required result is x*(4x^3 - 3x^2 +7) / (1- 2x^2)^2**