Multivariable chain rule to solve a third order derivative I came across this problem and I don't know how I should go about solving it:
"Express $\frac{\partial^3f}{\partial x^2\partial y}f(4x^2 + y, x + 1)$ in terms of partial derivatives of the function f"
I'm having a hard time grasping how these multivariable partial derivatives work. Earlier I had
$f(x, y) = g(u(x,y), v(x,y))$
For which i had to express
$\frac{\partial ^2f}{\partial x\partial y}$
In terms of partial derivatives of g, u and v and I got the following result:
$\frac{\partial ^2g}{\partial x\partial u} \frac{\partial ^2u}{\partial x\partial y} + \frac{\partial ^2g}{\partial x\partial v}\frac{\partial ^2v}{\partial x\partial y}$
But I'm very much doubting that this is correct, considering that I dont really understand this topic yet and my Calculus book doesn't seem to help me understand it.
 A: Define the auxiliary function $g(x,y)=f(4x^2+y,x+1)$ and let $u=4x^2+y$ and $v=x+1$. Then $g(x,y)=f(u,v)$ and the chain rule give
$$\frac{\partial g}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}=\frac{\partial f}{\partial u}(1)+\frac{\partial f}{\partial v}(0)=\frac{\partial f}{\partial u}$$
Then, \begin{align*}\frac{\partial^2g}{\partial x\partial y}&=\frac{\partial }{\partial x}\left(\frac{\partial g}{\partial y}\right)\\&=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial u}\right)\\&=\frac{\partial}{\partial u}\left(\frac{\partial f}{\partial u}\right)\frac{\partial u}{\partial x}+\frac{\partial}{\partial v}\left(\frac{\partial f}{\partial u}\right)\frac{\partial v}{\partial x}\\
&=\frac{\partial^2f}{\partial u^2}(8x)+\frac{\partial^2f}{\partial v\partial u}(1)\end{align*}
by the chain rule apply to $\frac{\partial f}{\partial u}$.
Then try to find
\begin{align*}
\frac{\partial^3g}{\partial x^2\partial y}&=\frac{\partial}{\partial x}\left(\frac{\partial^2g}{\partial x\partial y}\right)\\
&=\frac{\partial}{\partial x}\left(\frac{\partial^2f}{\partial u^2}(8x)+\frac{\partial^2f}{\partial v\partial u}(1)\right)
\end{align*}
And use chain rule to $\frac{\partial^2 f}{\partial u^2}$ and $\frac{\partial^2f}{\partial u\partial v}$ and the product rule.
I'll let you fill in the missing details.
