# Partial derivative reverse chain rule proof

I wish to know if I could do a reverse chain rule of partial derivative. By reverse I mean this:

$\frac{\partial z}{\partial y}$ $\div$ $\frac{\partial z}{\partial x}$ $=$ $\frac{\partial x}{\partial y}$

where $y=f(x)$ and $z=f(y)$

Is there a proof for this. I understand there is a proof for chain rule where I could state $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y} \times \frac{\partial y}{\partial x}$.

Is my first statement always true. Can I extend a proof for that based on the second statement (chain rule)

• Let $u(t) = f(g(t))$. Then $u'(t) = f'(g(t))g'(t)$. Let $v = u^{-1}$ (the inverse function) so that $v(u(t)) = t$. Then $v'(t) = \frac{1}{u'(v(t))}$. – reuns Aug 29 '17 at 12:32

Let $y=g(x)$ and $z=f(y)=f(g(x))$. Then, by the chain rule: $$\frac{\partial z}{\partial y} \div \frac{\partial z}{\partial x} =f'(g(x))\left[ \frac{\partial f}{\partial y} \frac{\partial y}{\partial x} \right]^{-1}=\left( \frac{\partial y}{\partial x} \right)^{-1} = g'(x)^{-1}$$ But, we can say a bit more, because: $$\frac{\partial }{\partial x} f^{-1}(x) = \left[ f'(f^{-1}(x)) \right]^{-1}$$ So then, since $x = g^{-1}(y)$, we get: $$\frac{\partial x}{\partial y} = \frac{\partial }{\partial y} g^{-1}(y)=\frac{1}{g'(g^{-1}(y))} = \frac{1}{g'(x)}=g'(x)^{-1}$$ as above. Thus, indeed: $$\frac{\partial z}{\partial y} \div \frac{\partial z}{\partial x}=\frac{\partial x}{\partial y}$$