simplifying second derivative

[vertex]

I'm following a tutorial on curve sketching and I got stuck on simplifying the second derivative. The answer they got is f" = [math]12x[/math]+[math]4[/math] / [math](x^2[/math] [math]-1)[/math] [math]^3[/math]
I'm not sure how they are getting that answer. Can someone explain?

 

[Deleted member 4993]

I'm following a tutorial on curve sketching and I got stuck on simplifying the second derivative. The answer they got is f" = [math]12x[/math]+[math]4[/math] / [math](x^2[/math] [math]-1)[/math] [math]^3[/math]
I'm not sure how they are getting that answer. Can someone explain?

From f'(x) → f"(x)

f"(x) = \(\displaystyle \frac{d}{dx} \left[ f'(x)\right] \ = \ \frac{d}{dx} \left[ \frac{-4*x}{(x^2 -1)^2} \right] \)............. apply quotient rule of differentiation

quotient rule of differentiation

\(\displaystyle \frac{d}{dx}\left[ \frac{u(x)}{v(x)} \right] \ = \ \frac{u' * v \ - \ u * v'}{v^2}\)

in this case:

u(x) = -4 * x

v(x) = (x^2 - 1)²

Continue......

 

[Cubist]

@vertex the work in your image is correct. However, please check that you didn't miss some parenthesis, and a square, when you quoted their answer. Should it be...

f''(x)= ( [MATH]12x\color{red}^2\color{black} + 4[/MATH] ) / [MATH](x^2-1)^3[/MATH] = [MATH]\frac{12x\color{red}^2\color{black} + 4}{\left(x^2 - 1\right)^3}[/MATH]
...because this would then be equivalent to your answer (after simplification).

If you have not made a mistake in quoting their answer, then your f'(x) is probably wrong. In this case, then please post the original question which hopefully includes a definition for f(x).