## Higher order differential equations and the chain rule (2 variables)

1. The problem statement, all variables and given/known data

The function F is deﬁned by F (r, θ) = f (x(r, θ), y(r, θ)), where f is twice continuously
diﬀerentiable and
x(r, θ) = r cos θ, y(r, θ) = r sin θ.
Use the chain rule to find

d2F/dθ2

2. Relevant equations

3. The attempt at a solution

I know that dF/dθ = (df/dx)(dx/dθ) + (df/dy)(dy/dθ)
and i can solve this

I know the next step is (d/dθ)(dF/dθ) but this is were i get lost. I must of missed the class this was explained in :(
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 Recognitions: Gold Member Science Advisor Staff Emeritus $$\frac{\partial F}{\partial \theta}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}$$ Since $x= r cos(\theta)$, $y= r sin(\theta)$, $x_\theta= -r sin(\theta)$ and $y_\theta= r cos(\theta)$ So $$\frac{\partial F}{\partial \theta}= - r sin(\theta)\frac{\partial f}{\partial x}+ r cos(\theta) \frac{\partial f}{\partial y}$$ which, I presume, is what you got. Now, $$\frac{\partial^2 F}{\partial \theta^2}= \frac{\partial}{\partial \theta}\left(\frac{\partial F}{\partial \theta}\right)$$ But the formula $$\frac{\partial \phi}{\partial \theta}= \frac{\partial \phi}{\partial x}\frac{\partial x}{\partial \theta}+ \frac{\partial \phi}{\partial y}\frac{\partial y}{\partial \theta}$$ is true for any function, $\phi$ and, in particular, for $$\phi= \frac{\partial F}{\partial \theta}= - r sin(\theta)\frac{\partial f}{\partial x}+ r cos(\theta) \frac{\partial f}{\partial y}$$ That is, $$\frac{\partial^2 F}{\partial \theta}= \frac{\partial}{\partial \theta}\left(-r sin(\theta)\frac{\partial f}{\partial x}+ r cos(\theta)\frac{\partial f}{\partial y}\right)$$ $$+ \frac{\partial}{\partial\theta}\left((-r sin(\theta)\frac{\partial f}{\partial x}+ r cos(\theta)\frac{\partial f}{\partial y}\right)$$ $$= - r cos(\theta)\frac{\partial f}{\partial x}- r sin(\theta)\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right)$$ $$- r sin(\theta)\frac{\partial f}{\partial y}+ r cos(\theta)\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial y}\right)$$ This is getting awkward to write as a single formula so just note that you do $$\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right)$$ and $$\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial y}\right)$$ using that same "chain rule formula": $$\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right)$$ $$= -r sin(\theta)\frac{\partial^2 f}{\partial x^2}+ r cos(\theta)\frac{\partial^2 f}{\partial x\partial y}$$ and $$\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial y}\right)$$ $$= r cos(\theta)\frac{\partial f}{\partial x\partial y}- r sin(\theta)\frac{\partial^2 f}{\partial y^2}$$
 Thanks for that, I know it must of taken a while to write out. I had a good look at it there now but some parts still dont make sense to me. I'll have another look at it tomorrow with fresh eyes :)
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