Partial Derivatives of sin(x^2 + y^2): Chain Rule

One chain rule, two partials

Write $u = x^2 + y^2$, so $f = \sin u$. The single variable chain rule does all the work. Differentiate the outer sine, then multiply by the partial derivative of $u$ with respect to whichever variable you are varying.

For $y$ the outer factor is identical. Only the inner partial changes, from $2x$ to $2y$, so compute $\cos(x^2+y^2)$ once and reuse it.

Where students lose the inner derivative

The usual error is writing $\frac{\partial f}{\partial x} = \cos(x^2 + y^2)$ and stopping. That is the derivative of $\sin u$ with respect to $u$, not with respect to $x$. It is off by exactly the factor $2x$. At $x = 0.7$, $y = 1.3$ the true partial is $-0.801$ and the truncated version gives $-0.572$, a ratio of $1.4$, which is $2x$.

A second slip is treating $y$ as constant so aggressively that $y^2$ disappears from inside the cosine. Holding $y$ fixed means its value stays put during the differentiation, not that it leaves the expression.

• Name the inner function before you differentiate anything.
• Write the outer derivative with $u$ still in it, then substitute back.
• Multiply by the inner partial as a separate written step, not mentally.

Reading the answer geometrically

Both partials carry the same cosine factor, so the gradient is $\cos(x^2+y^2)\,(2x,\, 2y)$, a scalar times the radial vector $(x,y)$. Wherever it is not the zero vector, the gradient points straight toward or straight away from the origin, which is what you expect from a surface depending only on distance from the origin.

It also tells you where both partials vanish at once: either $x = y = 0$, or $\cos(x^2+y^2) = 0$, meaning $x^2 + y^2 = \frac{\pi}{2},\ \frac{3\pi}{2},\ \dots$. Those circles are the crests and troughs of the ripple.