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I'm reading this.

It says,
from $ds^2 = dx^2 + dy^2$, we can get
$2dsd^2s =2dxd^2x $

I cannot understand this process.
Please give me some references.(simpler is better)

2 Answers 2

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If one feels uncomfortable with differentials it is possible to explain with ordinary derivatives.

Let's divide $ds^2 = dx^2 + dy^2$ by $dy^2$ to get $$ \left(\frac{ds}{dy}\right)^2=\left(\frac{dx}{dy}\right)^2+1\qquad\Leftrightarrow\qquad (s'(y))^2=(x'(y))^2+1. $$ Now differentiate both sides w.r.t. $y$ using the chain rule $$ 2s'\cdot s''=2x'\cdot x''\qquad\Leftrightarrow\qquad 2\frac{ds}{dy}\frac{d^2s}{dy^2}=2\frac{dx}{dy}\frac{d^2x}{dy^2}. $$ Multiplying by the denominator we obtain $2ds\cdot d^2s=2dx\cdot d^2x$

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It is just the chain rule with fixed $dy$, i.e $d (ds^2) = 2 ds \times d(ds) = 2 \hspace{2pt} ds \hspace{2pt} d^2 s$.

Similar for $dx$, and $d(dy) = 0$ as $dy$ is fixed.

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    forgive my ignorance but can you tell me wrt to which which variable they are differentiated ? – 2017-01-05
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    Here $d$ just means a small change in, they aren't being differentiated wrt to any variable (otherwise how would we know what $dx$ is, for example). – 2017-01-05
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    Oh that means the resultant statement means change of change ? – 2017-01-05
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    Yeh - think of acceleration for example – 2017-01-05
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    Thanks. last thing, When in calculus we learn these ? differential equation course ? – 2017-01-05
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    I can't remember explicitly being taught this, to be honest - it stems directly from the limit definition of derivatives – 2017-01-05
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    +1; Thanks for helping me, not many people on this site reply when asked questions on their answer by someone else. – 2017-01-05
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    No problem at all! – 2017-01-05