I have just started a course on differential equations, and unfortunetely enough for me, we immediately used notation foreign for me, for example:

$$ x^2 \left(\dfrac{d^2y}{dx^2}\right)^2 = \sin( x)\;\textrm{ and}\;\; y \times \dfrac{d^2 y}{dx^2} = \sin(x)$$ were used as examples of non-linear ordinary differential equations. My questions are

  • Is $\large \frac{dy}{dx}$ equal to $y'$? Also, what then is $\large\frac{d^2 y}{dx^2}\;$?

    I have looked up the formal mathematical definition but it somewhat confuses me,

  • Why on earth is this notation used, instead of just using $\;'\;$ or $\;''\;$? It seems to me much more confusing and unnecessarily messy.


Yes, your interpretation is correct: $$\frac{dy}{dx} = y' \;\text{ and likewise,}\;\;\frac{d^2y}{dx^2} = y''$$

It's often simpler to use $y'$, I agree. But there are contexts, for example in implicit differentiation, or when $x$ and $y$ are defined in terms of a parameter, like $t$, in which using $\frac{dy}{dx}$, e.g., makes explicit that we want to differentiate $y$ with respect to $x$. There are other advantages, but when no confusion or ambiguity results from using $y'$ to denote $\frac{dy}{dx}$, by all means, use it!

Please also see the previous thread Why the second derivative is written as $\dfrac{d^2y}{dx^2}$? for more comprehensive information as to what exactly the notation denotes, and its origin.

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  • $\begingroup$ Thank you. One final question; Why is $\dfrac{d^2 y}{dx^2}$ equal to $y''$? $\endgroup$ – Ylyk Coitus Feb 21 '13 at 20:40
  • $\begingroup$ It simply denoting the second derivative of y which respect to x, where you differentiate with respect to $x$ each time (twice). I do agree that the notation and placement of the "exponent" is confusing. I don't know if that clarifies anything! $\endgroup$ – amWhy Feb 21 '13 at 20:42
  • $\begingroup$ It still is somewhat fussy but I don't think it matters too much; I assume if you want the n-th derivative it just becomes $\dfrac{d^n y}{dx^n}$? $\endgroup$ – Ylyk Coitus Feb 21 '13 at 20:47
  • $\begingroup$ Yes it did! Thank you! $\endgroup$ – Ylyk Coitus Feb 21 '13 at 21:14

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