Hfrhyu

My book jumps from

$$fracpartial fpartial x(x, g(x)) + fracpartial fpartial y(x, g(x))g'(x) = 0 $$

to

$$fracpartial^2 fpartial x^2(x, g(x)) + 2 cdot fracpartial fpartial x partial y (x, g(x))g'(x) + fracpartial^2 fpartial y^2(x, g(x))(g'(x))^2 + fracpartial fpartial y(x, g(x))g''(x) = 0.$$

It is left as an exercise to verify that this new equality can be obtained by differentiating both sides of the first equation. I've been trying to do this, but I haven't been able to get to the desired result. I'm sort of new to partial derivatives, and I would really appreciate it if someone can show me the steps that are taken when differentiating the first equation. I'm pretty sure my setup itself is wrong.

I've looked at many examples now, but I still haven't been able to get anywhere, since they are not too similar to what I have

I would really appreciate any help.

Thanks

My try:

$$fracpartialpartial xleft(fracpartial fpartial x(x, g(x)) + fracpartial fpartial y(x, g(x))g'(x)right) $$

$$= underbracefracpartialpartial xleft(fracpartial fpartial x(x,g(x)) right)_textTerm 1 + underbracefracpartialpartial xleft(fracpartial fpartial y(x, g(x))g'(x)right)_textTerm 2$$

Now computing Term 1:

$$fracpartialpartial xleft(fracpartial fpartial x(x,g(x)) right) = fracpartial^2fpartial x^2(x, g(x)) cdot text some chain rule term $$

What would the chain rule term be? I know in single-variable calculus, if you're doing the derivative of $f(g(x))$, then you need to multiply by $g'(x)$. But here, there are two variables.

You're presumably given a function $f$ whose domain is a subset of $mathbb R^2$. Since this isn't specified, I'll take the simpler approach and assume the domain is $mathbb R^2$. This allow us to write $fcolon mathbb R^2 to mathbb R$.

Also presumably, you're given a differential function $gcolon mathbb Rto mathbb R$.

Now let's define $varphi$ as $varphi colon mathbb Rto mathbb R^2, xmapsto (x, g(x))$.

You want to differentiate the function given below (with the appropriate implicit domain)

$$
xmapsto fracpartial fpartial x(x, g(x)) + fracpartial fpartial y(x, g(x))g'(x)
$$

This can be rewritten as

$$
xmapsto left(fracpartial fpartial xcircvarphiright)(x) + left(fracpartial fpartial ycircvarphiright)(x)cdot g'(x)
$$

Since the derivative of a sum is the sum of the derivatives (when they all exist), we can focus on each member of this sum separately.

We want to differentiate $colorbluexmapsto left(dfracpartial fpartial xcircvarphiright)(x)$. Now allow me to change the notation $dfracpartial fpartial x$ to $partial_1f$. They denote the same thing: the derivative with respect to the first coordinate.

In the notation of this answer we have $m=2$, $n=1=p$, $G=partial_1f$, $F=varphi$, (now because the choice of letters in both questions gets confusing, I'll use $varphi_1$ and $varphi_2$ (in place of $f_1$ and $f_2$ used in the linked answer), $varphi_1colon mathbb Rto mathbb R, xmapsto x$ and $varphi_2colon mathbb Rto mathbb R, xmapsto g(x)$. Now set $H=Gcirc F$.

We have that $colorblueH$ is $colorblueleft(partial_1fright)circ varphi$ and this is what we wish to differentiate, in other words we wish to find $H'$. Since $H$ is a scalar function whose domain is $mathbb R$, its components are just a singular $h_1$ in the notation of the linked answer, which yields:

$$
beginalign
overbrace
beginbmatrix
partial _1h_1
endbmatrix_x
^H'(x)
&=
beginbmatrix
partial_1left(partial_1fright) & partial_2left(partial_1fright)
endbmatrix_varphi(x,y)
overbracebeginbmatrix
partial _1varphi_1\
partial_1varphi_2
endbmatrix_x^beginbmatrix
varphi_1'\
varphi_2'
endbmatrix_x\
&=
beginbmatrix
left(partial _1left(partial_1fright)right)(x,g(x)) & left(partial_2left(partial_1fright)right)(x, g(x))
endbmatrix
beginbmatrix
1\
g'(x)
endbmatrix\
&=
beginbmatrix
left(partial _1left(partial_1fright)right)(x,g(x)) + left(partial_2left(partial_1fright)right)(x, g(x))cdot g'(x)
endbmatrix_.
endalign
$$

The last term above can be translated back to

$$
dfracpartial^2fpartial x^2(x,g(x)) + dfracpartial^2fpartial ypartial x(x, g(x))cdot g'(x)
$$

It is similar for $dfracpartial fpartial ycircvarphi$.

Hint: The notation can be a bit confusing here. $f$ is a function of two variables. A derivative with respect to the first variable is denoted by $fracpartial fpartial x$, a derivative with respect to the second variable is denoted by $fracpartial fpartial y$. If instead you denote them by $fracpartial fpartial x_1$ and $fracpartial fpartial x_2$ it might be a bit clearer. At the same time you plug in the function $x$ into the first variable and the function $g(x)$ into the second variable. You are differentiating with respect to $x$ and need to apply the chain rule.

Edit: I will write out what happens in the first term with the renaming of variables. We want to compute: $$fracpartialpartial xleft(fracpartial fpartial x_1(x, g(x))right)$$
So we have
$$fracpartialpartial xleft(fracpartial fpartial x_1(x, g(x))right)=fracpartial partial x_1fracpartial fpartial x_1(x, g(x)) cdot fracpartial partial x(x) + fracpartial partial x_2fracpartial fpartial x_1(x, g(x)) cdot fracpartial partial xg(x)$$
Simplify a bit:

$$fracpartialpartial xleft(fracpartial fpartial x_1(x, g(x))right)=fracpartial^2 fpartial^2 x_1(x, g(x)) + fracpartial^2 fpartial x_2partial x_1(x, g(x)) cdot g'(x)$$
The second term is almost the same, you just have an extra factor of $g(x)$ that requires the product rule.