Find the partial derivative of ln(x^2 + y^2) with respect to x.

Step-by-step solution

1. 1

   Freeze $y$ as a constant. That is the whole meaning of a partial derivative with respect to $x$: every appearance of $y$ behaves like a fixed number, so $y^2$ differentiates to $0$.

2. 2

   Apply the chain rule for the logarithm. With $u=x^2+y^2$ and $\dfrac{d}{du}\ln u=\dfrac1u$:

   $\frac{\partial}{\partial x}\ln u=\frac{1}{u}\cdot\frac{\partial u}{\partial x}$

   Forgetting the inner derivative and writing $\tfrac{1}{x^2+y^2}$ alone is the standard error here.

3. 3

   Differentiate the inside with respect to $x$.

   $\frac{\partial}{\partial x}\left(x^2+y^2\right)=2x+0=2x$

4. 4

   Combine the two pieces.

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

   By the symmetry of the expression, the other partial is $\dfrac{\partial}{\partial y}=\dfrac{2y}{x^2+y^2}$ — simply swap the roles of $x$ and $y$.

5. 5

   Sanity-check the result. At $y=0$ the function reduces to $\ln x^2=2\ln|x|$, whose derivative is $\tfrac{2}{x}$; the formula gives $\tfrac{2x}{x^2}=\tfrac2x\;\checkmark$. The partial also vanishes all along the $y$-axis where $x=0$, matching the fact that $\ln\left(x^2+y^2\right)$ has a minimum in $x$ there for any fixed $y\ne 0$. The function is undefined at the origin, so the derivative is too.