7
$\begingroup$

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

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

$\endgroup$
1
  • 1
    $\begingroup$ Set $y=x^2$, what do you get? $\endgroup$ – Leon Sot Jan 3 '17 at 12:43
10
$\begingroup$

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

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

$\endgroup$
0
4
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ I like this answer better as it caters both x>0 and x<0 scenarios. $\endgroup$ – kevin Jan 3 '17 at 12:57
2
$\begingroup$

$$\large \frac{df(x)}{dg(x)}=\large \frac{\frac{df(x)}{dx}}{\frac{dg(x)}{dx}}=\frac{f'(x)}{g'(x)}$$ 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 : $$\frac{dx}{d(lnx)}=?\\=\frac{1}{\frac{1}{x}}=x$$

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.