Compute Derivative in Polar Coordinates

Question

I have the function $$\phi(r)=-192r^5+480r^4-440r^3$$ with $1/4\leq r\leq 3/4,$ in polar coordinates. So $r=\sqrt{x^2+y^2}.$ I want to compute $\phi'(r)$ and $\phi''(r).$ So, I use the chain rule, $$\frac{d\phi}{dr}=\frac{d\phi}{dx}\frac{dx}{dr}+\frac{d\phi}{dy}\frac{dy}{dr}$$ for the first derivative. And the same technique for the second derivative. I am right?

Answer 1

I found the solution. Because $r=r(x,y)=\sqrt{x^2+y^2}$ then $$\phi'(r)=\frac{d\phi}{dr}=\frac{d\phi}{dx}\frac{dx}{dr}+\frac{d\phi}{dy}\frac{dy}{dr}$$ and for the second derivative we must use chain rule again, so $$\phi''(r)=\frac{d^2\phi}{dr^2}=\frac{d}{dr}\left(\frac{d\phi}{dr}\right)=\frac{d}{dr}\left(\frac{d\phi}{dx}\frac{dx}{dr}+\frac{d\phi}{dy}\frac{dy}{dr}\right)=\cdots=$$ $$=\cdots=\phi_x\frac{d^2x}{dr^2}+\phi_y\frac{d^2y}{dr^2}+\phi_{xx}\left(\frac{dx}{dr}\right)^2+2\phi_{xy}\left(\frac{dx}{dr}\right)\left(\frac{dy}{dr}\right)+\phi_{yy}\left(\frac{dy}{dr}\right)^2$$

Answer 2

There is no need to bring in x or y nor domain of $r$.

Just differentiate $\phi$ two times w.r.t. $r.$