Multivariable calculus

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

Q: Can I use log rules to simplify before differentiating?

There is nothing to split here: \ln(x^{2}+y^{2}+1) is a log of a sum, and logs do not distribute over sums. Only a log of a product or a quotient can be broken up. Differentiate as it stands with the product and chain rules.

Q: Is the function defined everywhere?

Yes. The argument x^{2}+y^{2}+1 has minimum value 1 at the origin and grows outward, so it is always positive and the logarithm is always defined. The denominators in both partials are the same quantity, so they never vanish either.

For f(x,y) = x ln(x^2 + y^2 + 1), the partial with respect to x is ln(x^2 + y^2 + 1) + 2x^2/(x^2 + y^2 + 1), and the partial with respect to y is 2xy/(x^2 + y^2 + 1). The x-partial needs the product rule; the y-partial needs only the chain rule.

f(x,y) = x \ln\!\left(x^{2} + y^{2} + 1\right)

\frac{\partial f}{\partial x} = \ln\!\left(x^{2}+y^{2}+1\right) + \frac{2x^{2}}{x^{2}+y^{2}+1}

\frac{\partial f}{\partial y} = \frac{2xy}{x^{2}+y^{2}+1}

Why the +1 matters, and how to differentiate

The argument $x^{2}+y^{2}+1$ is at least 1 everywhere, so the logarithm is defined on the whole plane and the function is smooth. That is the point of the $+1$ : without it the log blows up at the origin and the partials would not exist there.

For $\frac{\partial f}{\partial x}$ , the function is a product of $x$ and $\ln(x^{2}+y^{2}+1)$ , and both factors depend on $x$ . Apply the product rule, then the chain rule with $\frac{d}{du}\ln u = \frac{1}{u}$ on the second piece.

\frac{\partial f}{\partial x} = \ln\!\left(x^{2}+y^{2}+1\right) + x \cdot \frac{2x}{x^{2}+y^{2}+1}

\frac{\partial f}{\partial x} = \ln\!\left(x^{2}+y^{2}+1\right) + \frac{2x^{2}}{x^{2}+y^{2}+1}

For $\frac{\partial f}{\partial y}$ the leading $x$ is a constant, so only the log is differentiated. The inside has $y$ -derivative $2y$ , and the log contributes the reciprocal of the inside.

\frac{\partial f}{\partial y} = \frac{2xy}{x^{2}+y^{2}+1}

The mistake: turning the log into a reciprocal too early

Students often write $\frac{\partial f}{\partial x} = \frac{x \cdot 2x}{x^{2}+y^{2}+1}$ and stop. That is the product rule with the first term missing. The derivative of $x$ is 1, and 1 times $\ln(x^{2}+y^{2}+1)$ is not zero, so the log term survives into the answer.

The $x$ -partial keeps a bare logarithm term. If your answer has no log in it, you dropped the first product rule term.
The $y$ -partial has no logarithm term, because the factor being differentiated is the log itself and the other factor is constant.
The chain rule numerator is $2x$ for the $x$ -partial and $2y$ for the $y$ -partial. Mixing them up is the other frequent slip.

One more check: swap $y$ for $-y$ in the original function and nothing changes, so $f$ is even in $y$ . An even function has an odd $y$ -partial, and $\frac{2xy}{x^{2}+y^{2}+1}$ does flip sign when $y$ does. Symmetry checks like this catch sign errors fast.

Evaluating on the x-axis

At $(1,0)$ the inside is $2$ , so $\frac{\partial f}{\partial x}(1,0) = \ln 2 + \frac{2}{2} = \ln 2 + 1$ , roughly $1.693$ .

\nabla f(1,0) = \langle \ln 2 + 1,\; 0 \rangle

The $y$ -component is zero along the entire $x$ -axis, since every term in $\frac{2xy}{x^{2}+y^{2}+1}$ carries a factor of $y$ . Geometrically the $x$ -axis is a ridge line of the surface in the $y$ direction.

Frequently asked questions

Can I use log rules to simplify before differentiating?

There is nothing to split here: $\ln(x^{2}+y^{2}+1)$ is a log of a sum, and logs do not distribute over sums. Only a log of a product or a quotient can be broken up. Differentiate as it stands with the product and chain rules.

Is the function defined everywhere?

Yes. The argument $x^{2}+y^{2}+1$ has minimum value 1 at the origin and grows outward, so it is always positive and the logarithm is always defined. The denominators in both partials are the same quantity, so they never vanish either.