Multivariable Chain Rule. Where is my mistake?

Question

Say $x=\sqrt{z^2+y^2}$.

For a function $f(y,z)$, we have

$$ \tag{1} \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x} \ . $$

By definition of $x$, we have \begin{align} \frac{\partial y}{\partial x} = \frac{x}{\sqrt{x^2 - z^2}} = \frac{x}{y} \ , \tag{2}\\ \frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^2 - y^2}} = \frac{x}{z} \ .\tag{3}\\ \end{align} Therefore $$ \tag{4} \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}\frac{x}{y} + \frac{\partial f}{\partial z}\frac{x}{z} \ . $$

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Now assume $f(y,z) = e^{\sqrt{y^2 + z^2}}$. We have

\begin{align} \frac{\partial f}{\partial y} = e^{\sqrt{y^2 + z^2}} \frac{y}{\sqrt{y^2 + z^2}} = e^x\frac{y}{x}\ , \tag{5}\\ \frac{\partial f}{\partial z} = e^{\sqrt{y^2 + z^2}} \frac{z}{\sqrt{y^2 + z^2}} = e^x\frac{z}{x}\ . \tag{6}\\ \end{align} By $(4)$, we have $$ \tag{7} \frac{\partial f}{\partial x} = e^x\frac{y}{x}\frac{x}{y} + e^x\frac{x}{z}\frac{z}{x} = 2e^x\ . $$ But that can't be right, because $f = e^x$, so it must be $$ \tag{8} \frac{\partial f}{\partial x} = e^x \neq 2e^x \ . $$

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So where is my mistake?

Answer 1

What you've written really doesn't make sense. You start with a function $f(u)=e^u$ — a function of a single variable. Then you have $u=g(y,z)$. So, letting $F(y,z) = f(g(y,z))$, we have $$\frac{\partial F}{\partial y} = f'(g(y,z))\frac{\partial g}{\partial y},$$ and similarly for the $z$ partial.

With $f(u) = e^{\sqrt u}$, this gives $$\frac{\partial F}{\partial y} = \frac1{2\sqrt{y^2+z^2}}e^{\sqrt{y^2+z^2}}\cdot 2y = \frac y{\sqrt{y^2+z^2}}e^{\sqrt{y^2+z^2}}.$$