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I'm studying for my exam Math and I came across a problem with one exercise.

Calculate the second derivative of $$f'(x)=\frac{2x^3-3x^2}{(x^2-1)^2}$$

I just can't seem to calculate the second derivative of this rational function. If someone could help me . I just can't seem to get the right answer

See my calculations and what the right answer should be according to my teacher

enter image description here

Thanks!

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  • $\begingroup$ Maybe to help you structure the formula, try to work with a product instead of a fraction:$(2x^3-3x^2)(x^2-1)^{-2}$. $\endgroup$ – Raskolnikov Jun 10 '18 at 11:06
  • $\begingroup$ Differentiating $u/v$ gives $u'/v-uv'/v^2$; differentiating again gives $u''/v-2u'v'/v^2-uv''/v^2+2uv'^2/v^3$. $\endgroup$ – J.G. Aug 9 '18 at 5:48
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In your very first step, where you have $6x^2 - 6x \cdot (x^2 - 1)^2 \ldots$ in magenta, the first part -- the $6x^2 - 6x$ -- should be in parentheses.

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I split the fraction into two parts, then used the product rule on both parts.

$\dfrac{2x^3-3x^2}{(x^2-1)^2} = (2x^3*(x^2-1)^{-2}) - (3x^2*(x^2-1)^{-2})$

After product rule on both terms we have

$\dfrac{12x^3-8x^4}{(x^2-1)^3} + \dfrac{6x^2-6x}{(x^2-1)^2}$

$\dfrac{12x^3-8x^4}{(x^2-1)^3} + \dfrac{(6x^2-6x)*(x^2-1)}{(x^2-1)^{2}*(x^2-1)}$

After simplifying the numerators and factoring out a $-2x$, we are left with

$f''(x)=\dfrac{-2x(x^3-3x^2+3x-3)}{(x^2-1)^{3}}$

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