Second Order Partial Derivatives + Chain Rule

Homework Statement

Let z = z (x,y) be a function with x = x(s), y = y(t) satisfying the partial differential equation

(Ill write ddz/ddt for the partial derivative of z wrt t and
dz/dt for the total derivative of z wrt t, as I have no idea how to use Latex.)

ddz/ddt + 1/2s^2(dd^2z/dds^2) + s(ddz/dds) - z = 0

Using chain rule, show that z satisfies

-2z + 2(dy/dt)(ddz/ddy) + s((2dx/ds + s(d^2x/ds^2))ddz/ddx + s(dx/ds)^2 dd^2z/ddx^2) = 0

The Attempt at a Solution

So I wrote out my little "tree" with the variables and found

ddz/ddt = ddz/ddy * dy/dt

ddz/dds = ddz/ddx * dx/ds

Where I think im struggling is

dd^2/dds^2 = dd/dds[ddz/ddx*dx/ds]

Im not too sure how I should be computing this derivative and all attempts have given errors so far.

Im pretty sure all I need to do is work this derivative out then sub everything back into my original equation, tidy it up a little and then I get my desired result.

Anyone steer me in the right direction please.

SammyS
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Let z = z (x,y) be a function with x = x(s), y = y(t) satisfying the partial differential equation

$$\frac{\partial z}{\partial t}+\frac{1}{2}\,s^2\,\frac{\partial^2z}{\partial s^2}+s\,\frac{\partial z}{\partial s }-z=0$$

Using chain rule, show that z satisfies

-2z + 2(dy/dt)(ddz/ddy) + s((2dx/ds + s(d^2x/ds^2))ddz/ddx + s(dx/ds)^2 dd^2z/ddx^2) = 0

-2z + 2(dy/dt)(∂z/∂y) + s((2dx/ds + s(d2x/ds2))∂z/∂x + s(dx/ds)^2 ∂2z/∂x2) = 0

The Attempt at a Solution

...

2z/∂s2 = ∂/∂s[∂z/∂x*dx/ds]
...
Now try the product rule, then the chain rule. (Remember that ∂z/∂x is a function of x(s).)

$$\frac{\partial}{\partial s}\left(\frac{\partial z}{\partial x}\,\cdot\,\frac{dx}{ds}\right) =\left(\frac{\partial}{\partial s}\,\frac{\partial z}{\partial x}\right)\cdot\,\frac{dx}{ds}+\frac{\partial z}{\partial x}\,\cdot\,\frac{d^2x}{ds^2}$$

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