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I need help finding the second derivative of this function. I found the first derivative and the second, but the program says my answer is incorrect either by typing error and it won't process correctly or the answer is plain out wrong.The problem is I don't know which so have I gone about solving this problem correctly? And are the first and second derivatives correct? If not, a step by step explanation would be greatly appreciated. Thanks in advance $$f(x)=\frac{cos(x)}{(2x)}$$
so $$f'(x)=\frac{(2x)(-sinx)-[(cosx)(2)]}{(2x)^2}$$ and $$f''(x)=\frac{(4x^2)((2x)(-cosx))-[((2x)(-sinx)-2cosx)(8x)]}{(2x)^4}$$

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  • $\begingroup$ $[(2x)(−\sin{x})]'=-2\sin{x}-2x\cos{x}$ $\endgroup$ – M. Strochyk Jun 25 '14 at 20:40
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The second derivative of a product is given by $$(UV)''=U''V+2U'V'+V''.$$

Here, $U=\cos x$ ($\rightarrow -\sin x\rightarrow-\cos x$) and $V=\frac1x$ ($\rightarrow -\frac1{x^2}\rightarrow\frac2{x^3}$), so that

$$f''(x)=\frac12(-\frac{\cos x}x+2\frac{\sin x}{x^2}+2\frac{\cos x}{x^3}).$$

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  • $\begingroup$ In the same vein, $$\Big(\frac UV\Big)'=\frac {U''}V-2\frac{U'V'}{V^2}-\frac{UV''}{V^2}+2\frac{U(V')^2}{V^3}=\frac{U''V^2-2U'VV'-UVV''+2U(V')^2}{V^3}.$$ $\endgroup$ – Yves Daoust Jun 26 '14 at 6:24
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The first derivative is correct, although it simplifies to $-2\frac{x\sin x + \cos x}{4x^2} = -\frac{x\sin x + \cos x}{2x^2}$. Your calculation for the second derivative is not correct --- it's easier to do it correctly if you have simplified the numerator. In the first term, you failed to completely take the derivative of the numerator of $f'(x)$. The correct answer is (starting from the simplified form $-\frac{x\sin x + \cos x}{2x^2}$) \begin{align} f''(x) &= -\frac{(2x^2)(\sin x + x\cos x - \sin x) - (x\sin x + \cos x)(4x)}{(2x^2)^2} \\ &= -\frac{2x^3\cos x - 4x^2\sin x - 4x\cos x}{4x^4} \\ &= -\frac{2x^2\cos x - 4x\sin x - 4\cos x}{4x^3} \\ &= \frac{-x^2\cos x + 2x\sin x + 2\cos x}{2x^3}. \end{align}

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  • $\begingroup$ @amWhy Is that comment directed at my answer? 'Cause I think it's right... $\endgroup$ – rogerl Jun 25 '14 at 20:51
  • $\begingroup$ @amWhy No, I just started from a simplified version of $f'$. I updated my answer. Thanks. $\endgroup$ – rogerl Jun 25 '14 at 20:56
  • $\begingroup$ Very good. I was working with the OPs derivative, so your first line confused me. Glad you updated your post to show your simplification of the first derivative. $\endgroup$ – Namaste Jun 25 '14 at 20:58

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