To see how this should work, consider the first derivative.
There, the formula is
d*y*/d*x* = (d*y*/d*t*)/(d*x*/d*t*).
That is, simply divide both sides of the fraction by d*t*.
Another even slicker way to do this
would be to reinterpret the differentials
as derivatives with respect to *t*;
that is, writing a dot above a quantity
to indicate differentiation with respect to *t*,
write d*y*/d*x* = *ẏ*/*ẋ*.

If you try to do this with second derivatives,
based on the usual notation for them,
then you get a formula which is *wrong*:
d^{2}*y*/d*x*^{2} ≠
*ÿ*/*ẋ*^{2} =
(d^{2}*y*/d*t*^{2})/(d*x*/d*t*)^{2}.
(Here, I've written ‘≠’
to show that ‘=’ would have been wrong,
but it's possible that these may happen to be equal in certain examples.)
To get the correct formula,
we simply need to differentiate *ẏ*/*ẋ*
using the Quotient Rule:
(d/d*t*)(*ẏ*/*ẋ*) =
(*ẋ**ÿ* −
*ẏ**ẍ*)/*ẋ*^{2}.
Dividing by *ẋ*,
the second derivative of *y* with respect to *x* is really
(*ẋ**ÿ* −
*ẏ**ẍ*)/*ẋ*^{3} =
*ÿ*/*ẋ*^{2} −
*ẏ*/*ẋ* ⋅
*ẍ*/*ẋ*^{2};
in other words, the naïve formula
is only the first term of a two-term expression.
This formula is a little long,
but it will correctly give you
the second derivative of *y* with respect to *x*
using the first and second derivatives of *x* and *y*
with respect to *t*.

There is a symbol for the second derivative using differentials
that can serve as a mnemonic for this.
To get it, we again differentiate d*y*/d*x* using the Quotient Rule,
only now using the Quotient Rule for differentials
rather than the Quotient Rule for derivatives:
d(d*y*/d*x*) =
(d*x* d(d*y*) −
d*y* d(d*x*))/(d*x*)^{2} =
(d*x* d^{2}*y* −
d*y* d^{2}*x*)/d*x*^{2}.
Dividing by d*x*,
the second derivative of *y* with respect to *x* is really
(d*x* d^{2}*y* −
d*y* d^{2}*x*)/d*x*^{3} =
d^{2}*y*/d*x*^{2} −
d*y*/d*x* ⋅ d^{2}*x*/d*x*^{2}.
As you can see, replacing d with d/d*t*
gives the formula from the previous paragraph.

For this reason,
I don't like to write d^{2}*y*/d*x*^{2}
for the second derivative of *y* with respect to *x*.
Of course, nobody wants to write the formula from the previous paragraph
when they just want a symbol for the second derivative;
fortunately, you can write (d/d*x*)^{2}*y* for that.
This simply means that you apply the operation d/d*x*
(find the differential and then divide by d*x*,
or equivalently find the derivative with respect to *x*)
twice
to get the second derivative,
which is certainly correct.
You can even use this as a mnemonic for finding this second derivative:
instead of interpeting d/d*x*
as taking the differential and then dividing by d*x*,
interpret it as taking the derivative with respect to *t*
and then dividing by *ẋ*.
This is essentially how the book tells you to take the second derivative.

Finally, whether you use either
(d*x* d^{2}*y* −
d*y* d^{2}*x*)/d*x*^{3}
or (d/d*x*)^{2}*y*,
either way you can perform practical caclulations
by interpreting the differentials literally as differentials.
You simply have to write everything in terms of *t*,
put d*t* and d^{2}*t* in where they naturally appear,
and find that the differentials of *t* cancel in the final answer.
Alternatively, anticipating that the differentials of *t* will cancel,
you can ignore them,
which turns taking differentials
into taking derivatives with respect to *t* again.

I'll do Example 10.2.2 on page 571 of the textbook
to illustrate all of these approaches.
Given *x* = *t* − *t*^{2},
d*x* = d*t* − 2*t* d*t*,
or *ẋ* = d*x*/d*t* = 1 − 2*t*.
Next, d^{2}*x* =
d^{2}*t* −
2 d*t*^{2} −
2*t* d^{2}*t*
(applying the Product Rule in the second term),
while *ẍ* = −2.
Similarly, given *y* = *t* − *t*^{3},
d*y* = d*t* − 3*t*^{2} d*t*,
or *ẏ* = 1 − 3*t*^{2}.
Next, d^{2}*y* =
d^{2}*t* −
6*t* d*t*^{2} −
3*t*^{2} d^{2}*t*,
while *ÿ* = −6*t*.

To find (d/d*x*)*y* = d*y*/d*x*,
either directly divide
(d*t* −
3*t*^{2} d*t*)/(d*t* −
2*t* d*t*)
and simplify this (by cancelling factors of d*t*)
to (1 − 3*t*^{2})/(1 − 2*t*),
or instead divide *ẏ*/*ẋ*,
which again gives
(1 − 3*t*^{2})/(1 − 2*t*).
This is pretty much the same process,
no matter how you go about it.
Then to find (d/d*x*)^{2}*y*,
one way is to note that
(d/d*x*)*y* =
(1 − 3*t*^{2})/(1 − 2*t*)
from the previous paragraph,
so differentiate with respect to *x* again.
Either take
d((1 − 3*t*^{2})/(1 − 2*t*)) =
(2 d*t* − 6*t* d*t* +
6*t*^{2} d*t*)/(1 − 2*t*)^{2}
divided by d*x* = d*t* − 2*t* d*t*
and simplify by cancelling factors of d*t*,
or take
(d/d*t*)((1 −
3*t*^{2})/(1 − 2*t*)) =
(2 − 6*t* +
6*t*^{2})/(1 − 2*t*)^{2}
divided by *ẋ* = 1 − 2*t*;
either way, you get
(d/d*x*)^{2}*y* =
(2 − 6*t* + 6*t*^{2})/(1 −
2*t*)^{3}.
This is essentially how the textbook does this problem.

Alternatively, using (d/d*x*)^{2}*y* =
(d*x* d^{2}*y* −
d*y* d^{2}*x*)/d*x*^{3},
we immediately get
((d*t* −
2*t* d*t*)(d^{2}*t* −
6*t* d*t*^{2} −
3*t*^{2} d^{2}*t*) −
(d*t* −
3*t*^{2} d*t*)(d^{2}*t* −
2 d*t*^{2} −
2*t* d^{2}*t*))/(d*t* −
2*t* d*t*)^{3},
which simplifies drastically to
(2 − 6*t* +
6*t*^{2})/(1 − 2*t*)^{3},
the same answer as above.
Notice that there is no need to work out d*y*/d*x* first.
Or using (d/d*x*)^{2}*y* =
(*ẋ**ÿ* −
*ẏ**ẍ*)/*ẋ*^{3},
you immediately get
((1 − 2*t*)(−6*t*) −
(1 − 3*t*^{2})(−2))/(1 −
2*t*)^{3},
which simplifies (somewhat less drastically)
to (2 − 6*t* +
6*t*^{2})/(1 − 2*t*)^{3}
again.
I prefer this last method,
which gets the answer is one step after the preliminary calculations
and doesn't require quite as much algebra to simplify
as the corresponding method using differentials.

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This web page was written in 2016 by Toby Bartels, last edited on 2016 February 21. Toby reserves no legal rights to it.

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