First and second derivative of x^2 - x / x^3 - 4x^2 - x +4?

Question

First and second derivative of x^2 - x  /  x^3 - 4x^2 - x +4?

anonymous · Answer

For the first derivative, I got

anonymous · Answer

f'(x) = -x^4 + 2x^3 - 5x^2 + 8x - 4  /

anonymous · Answer

(x^3 - 4x^2 - x + 4)^2

anonymous · Answer

But for the second derivative, I'm getting a giant expression for the numerator... I need help

anonymous · Answer

First thing I would do is factor the numerator and denominator and cancel out common factors. This makes the differentiation using quotient rule easier.

anonymous · Answer

Okay I will try that

anonymous · Answer

Do you mean by (x-1)?

anonymous · Answer

Wait, were you talking about finding the second derivative?

anonymous · Answer

$$\frac{ x ^{2}-x }{ x ^{3}-4x ^{2}-x+4 }=\frac{ x(x-1) }{(x ^{3}-4x ^{2})-(x-4) }=\frac{ x(x-1) }{ x ^{2}(x-4)-1(x-4)}=\frac{ x(x-1) }{ (x ^{2}-1)(x-4) }=\frac{ x(x-1) }{(x+1)(x-1)(x-4)  }$$.
I would find the first derivative after factoring then find the second from the first

the_fizicx99 · Answer

@hullsnipe LaTeX isn't working >.< niether is the drawing tool

the_fizicx99 · Answer

Neither*

anonymous · Answer

I tried to use the equation tool too, but it's not working :(

anonymous · Answer

Thanks hullsnipe, but I don't think I'd be able to recognize that...