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Skill 10 of 14
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Deeper derivatives

When speed $v=s’$ itself is changing, $v’=s^{\prime\prime}$, we reach a deeper derivative. In Leibniz notation:

$\displaystyle v’=\frac{\mathrm dv}{ \mathrm dt}=\frac{ \mathrm d\overbrace{( \mathrm ds/ \mathrm dt)}^v}{ \mathrm dt}=$1$ \displaystyle \frac{ \mathrm d\,( \mathrm ds)/ \mathrm dt}{ \mathrm dt}=\frac{ \mathrm d\,( \mathrm ds)}{ \mathrm dt\, \mathrm dt} =\frac{ \mathrm d\,( \mathrm ds)}{ \mathrm dt^2} =$2$\displaystyle\frac{ \mathrm d^2s}{ \mathrm dt^2}=s^{\prime\prime}$

The notation $\frac{\mathrm d^2s}{\mathrm dt^2}$ is thus a “signal” that a double derivative is taking place, just like $s^{\prime\prime}$ is. Should there be more changes of changes of changes etc., then we’ll just continue this pattern:

$$\frac{\mathrm ds}{\mathrm dt}\qquad
\frac{\mathrm d^2s}{\mathrm dt^2}\qquad
\frac{\mathrm d^3s}{\mathrm dt^3}\qquad
\frac{\mathrm d^4s}{\mathrm dt^4}\qquad
\frac{\mathrm d^5s}{\mathrm dt^5}\qquad \cdots$$ $$s^{\prime}\qquad
s^{\prime \prime }\qquad
s^{\prime \prime \prime }\qquad
s^{\prime \prime \prime \prime }\qquad
s^{\prime \prime \prime \prime \prime } \qquad \cdots$$

Note that sometimes the involved property is placed down in front:

$$\frac{\mathrm ds}{ \mathrm dt}= \frac{\mathrm d}{ \mathrm dt}s$$

This might not feel strictly, mathematically correct – it looks as if $\frac{\mathrm d}{\mathrm dt}$ is treated as merely a symbol – but this is a notation form, you can meet now and then.

You may be asking now, why this Leibniz notation, $\mathrm ds/ \mathrm dt$, is necessary when we can just use the prime notation, $s’$. The reason is that we might at some point want to go “behind the scenes” and work with or rearrange the involved terms; terms such as the time step $\mathrm dt$. When this is not needed, we can stick to the prime notation and just be ready to change it to Leibniz notation on the fly, should it become necessary.

Finally note: if there are many primes, such as $s^{\prime \prime \prime \prime \prime \prime }$, it is not easy to read it. You might then instead see people write a raised number in a bracket: $s^{ \prime \prime \prime \prime \prime \prime }=s^{(6)}$. This of course looks dangerously similar to an exponent, $s^6$ (which means $s^6=s\cdot s \cdot s \cdot s \cdot s \cdot s $, so something very different), so you might want to stick to Leibniz notation instead.