What is the second derivative of $$f(x) = \frac{4x}{x^2-4}?$$ I have tried to use the quotient rule but I can't seem to get the answer.
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Have you found the first derivative, yet? – Jun 12 '16 at 01:28
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Formatting tips here. – Em. Jun 12 '16 at 01:30
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It's tedious, but if you know the quotient rule, you can find it – JasonM Jun 12 '16 at 01:32
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Also, the answer in the back of the book or as your teacher wrote it may look different, so it's important that you show your answer is equivalent. – JasonM Jun 12 '16 at 01:33
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It should be easier to express your function as $2/(x+2)+2/(x-2)$... – Jun 12 '16 at 01:35
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To simplify the differentiation, we can first rewrite the function as a sum of partial fractions. To do this, we assert that $\frac{4x}{x^2 - 4}$ can be written as $\frac{A}{x + 2} + \frac{B}{x - 2},$ with $A$ and $B$ constants. If this is the case, then clearly, $(A + B)x + 2(B - A) = 4x,$ from which we conclude that $A = B = 2.$ The decomposition is thus $\frac{2}{x + 2} + \frac{2}{x - 2}.$
Let us differentiate this expression twice. Notice that the first derivative can be computed as $$\frac{d}{dx}\left[\frac{2}{x + 2} + \frac{2}{x - 2}\right]$$ $$= -2(x + 2)^{-2} - 2(x - 2)^{-2}.$$
The second derivative, likewise, is easy to compute: $$f''(x) = \boxed{4(x + 2)^{-3} + 4(x - 2)^{-3}}.$$
And we are done! Hope this helped!
K. Jiang
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Avoiding the quotient rule, just for an option: $$\begin{align} \ln(f(x)) &=\ln(4)+\ln(x)-\ln(x+2)-\ln(x-2)\\ \frac{f'(x)}{f(x)} &=x^{-1}-(x+2)^{-1}-(x-2)^{-1}\\ f'(x) &=f(x)\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)\\ f''(x) &=f(x)\left(-x^{-2}+(x+2)^{-2}+(x-2)^{-2}\right)+f'(x)\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)\\ &=f(x)\left(-x^{-2}+(x+2)^{-2}+(x-2)^{-2}\right)+f(x)\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)^2\\ &=f(x)\left[-x^{-2}+(x+2)^{-2}+(x-2)^{-2}+\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)^2\right]\\ &=\frac{4x}{x^2-4}\left[-x^{-2}+(x+2)^{-2}+(x-2)^{-2}+\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)^2\right]\\ \end{align}$$
2'5 9'2
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@K.Jiang Partial fractions does the job better, imho. Just wanted to put up logarithmic differentiation as an option. – 2'5 9'2 Jun 12 '16 at 02:09
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Yes, logarithmic differentiation is almost always a clever approach. Obviously here, differentiating by quotient rule and brute force would not be wise :) – K. Jiang Jun 12 '16 at 02:12
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As Jonh said, $f(x)=\frac{2}{x+2}+\frac{2}{x-2}$, and so,
$f'(x)=\frac{-2}{(x+2)^2}+\frac{-2}{(x-2)^2}$, and
$f''(x)=\frac{4}{(x+2)^3}+\frac{4}{(x-2)^3}$.
Majid
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Check your calculation for the second derivative; it seems to have incorrect signs. – K. Jiang Jun 12 '16 at 01:48
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