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.
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.