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$$\frac{d}{dx^2}x=\frac{1}{2x}$$

I can't wrap my head around that. It appeared in a physics context, i.e. it may not be mathematically rigorous.

Thanks for some explanation.

Wolfram Alpha tells me I cannot differentiate by $x^2$: Invalid value

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    $\begingroup$ Set $y=x^2$, what do you get? $\endgroup$
    – Leon Sot
    Jan 3, 2017 at 12:43

4 Answers 4

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Call your variable $\;t:=x^2\implies \sqrt t=x\;,\;\;t\ge0\;$ , so you want

$$\frac d{dx^2}(x)=\frac d{dt}(\sqrt t)=\frac1{2\sqrt t}=\frac1{2x}$$

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Let $f(x^2)=x$.

Hence, for $x>0$ $$f'(x^2)=(\sqrt{x^2})'_{x^2}=\frac{1}{2\sqrt{x^2}}=\frac{1}{2x}$$

If $x<0$ so $$f'(x^2)=-(\sqrt{x^2})'_{x^2}=-\frac{1}{2\sqrt{x^2}}=\frac{1}{2x}$$

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    $\begingroup$ I like this answer better as it caters both x>0 and x<0 scenarios. $\endgroup$
    – kevin
    Jan 3, 2017 at 12:57
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$$\large \frac{df(x)}{dg(x)}=\large \frac{\frac{df(x)}{dx}}{\frac{dg(x)}{dx}}=\frac{f'(x)}{g'(x)}$$ for example $$f(x)=x^4+x^2+1 ,g(x)=x^2 $$ $$\frac{df}{dg}=\frac{d(x^4+x^2+1)}{d(x^2)}=\frac{4x^3+2x}{2x}=2x^2+1$$ if you take $u=x^2$ you need to find $\frac{df}{du}$ so first rewrite $f(x)$ by $u$ $$f(x)=(x^2)^2+x^2+1=u^2+u+1 \\ \to \frac{df}{du}=2u+1=2x^2+1$$ another example :$$f(x)=sinx ,g(x)=tan x \\\frac{df}{dg}=\frac{cos x}{tan^2 x+1}=sec^3x$$

another example : $$\frac{dx}{d(lnx)}=?\\=\frac{1}{\frac{1}{x}}=x$$

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$$\frac{d(x)}{dx} \cdot \frac{dx}{dx^2}=1 \cdot \frac{1}{dx^2/{dx}}=\frac{1}{2x}.$$

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