Partial Derivatives of e^x ln(1 + y^2)

A separated product needs no product rule

This function has the form $g(x)h(y)$, with $g(x) = e^{x}$ and $h(y) = \ln(1+y^{2})$. When $x$ moves, the whole factor $h(y)$ is a constant multiplier, so you differentiate $g$ and leave $h$ alone.

When $y$ moves, $e^{x}$ is the constant multiplier and you differentiate the logarithm with the inside derivative $2y$ over the inside $1+y^{2}$.

At $(0,1)$ the slopes are $\ln 2 \approx 0.693$ in $x$ and exactly $1$ in $y$, since $\frac{2\cdot 1}{1+1} = 1$.

The mistake: reaching for the product rule

• Writing $\frac{\partial f}{\partial x} = e^{x}\ln(1+y^{2}) + e^{x}\cdot\frac{2y}{1+y^{2}}$ applies the product rule as if both factors depended on $x$. The second factor does not, and its $x$ derivative is $0$, so that extra term is pure error.
• Writing $\frac{\partial f}{\partial y} = \frac{e^{x}}{1+y^{2}}$ forgets the inside derivative $2y$.
• Writing $\frac{\partial f}{\partial y} = e^{x}\ln(2y)$ mangles the logarithm rule entirely. Differentiating $\ln(u)$ never leaves a logarithm behind.

The line $y = 0$ is a good test. There $\ln(1+y^{2})$ sits at its minimum, so the slope in $y$ must be $0$ for every $x$. The correct formula has the factor $2y$ in the numerator and gives exactly that; the version missing it does not.

The x partial is the function again

Because $e^{x}$ is its own derivative and $x$ appears nowhere else, this function satisfies $\frac{\partial f}{\partial x} = f$. Every horizontal line is a scaled copy of the exponential, so moving one unit in $x$ multiplies the height by $e$, whatever $y$ is.

The $y$ behaviour is different in character. The factor $\ln(1+y^{2})$ is zero on the $x$-axis, rises in both directions, and grows only logarithmically, so the slope $\frac{2y}{1+y^{2}}$ fades back toward $0$ once $|y|$ is large.