Multivariable calculus

Gradient of e^(xy) sin(z) in Three Variables

Q: Why do the x and y components look identical apart from the swap?

Because f is symmetric in x and y: the only place they appear is in the product xy. Swapping them leaves the function unchanged, so it must swap the two partials as well, and y e^{xy}\sin z becomes x e^{xy}\sin z.

Q: Does the gradient ever point purely in the z direction?

Yes, everywhere on the planes z = k\pi, where \sin z = 0 kills the first two components while \cos z = \pm 1 keeps the third alive. It also happens on the z axis, where x = y = 0 makes the first two components vanish, apart from the critical points z = \frac{\pi}{2}+k\pi, where the whole vector is zero and no direction is defined.

For f(x, y, z) = e^(xy) sin(z) the gradient is (y e^(xy) sin z, x e^(xy) sin z, e^(xy) cos z). For the x and y partials, sin z is a constant multiplier and the chain rule acts on the exponent xy. For the z partial, e^(xy) is the constant. The gradient vanishes only where x = y = 0 and cos z = 0.

f(x,y,z) = e^{xy}\sin z

\nabla f = \left\langle y\,e^{xy}\sin z,\; x\,e^{xy}\sin z,\; e^{xy}\cos z \right\rangle

\frac{\partial f}{\partial x} = y\,e^{xy}\sin z

\frac{\partial f}{\partial y} = x\,e^{xy}\sin z

\frac{\partial f}{\partial z} = e^{xy}\cos z

Decide which factor is frozen, then differentiate the other

The function splits into a factor in $x$ and $y$ and a factor in $z$ alone. For the first two components, $\sin z$ is a constant multiplier and all the work is the chain rule on the exponent.

\frac{\partial f}{\partial x} = \sin z \cdot e^{xy}\cdot\frac{\partial}{\partial x}(xy) = y\,e^{xy}\sin z

\frac{\partial f}{\partial y} = \sin z \cdot e^{xy}\cdot\frac{\partial}{\partial y}(xy) = x\,e^{xy}\sin z

For the third component the roles reverse. Nothing in $e^{xy}$ mentions $z$ , so it comes along as a multiplier and the sine is differentiated.

\frac{\partial f}{\partial z} = e^{xy}\cdot\frac{d}{dz}\left(\sin z\right) = e^{xy}\cos z

Every component keeps the full $e^{xy}$ . It is a factor of the original function, and no partial derivative here can remove it, since the exponential is never differentiated away.

Where the gradient is zero, and a point check

The factor $e^{xy}$ is positive everywhere, so it can be divided out of all three equations. What remains is $y\sin z = 0$ , $x\sin z = 0$ and $\cos z = 0$ .

The third equation forces $z = \frac{\pi}{2}+k\pi$ , and at those values $\sin z = \pm 1$ , which is not zero. So the first two equations reduce to $y = 0$ and $x = 0$ .

\nabla f = \left\langle 0,0,0 \right\rangle \iff x = 0,\; y = 0,\; z = \tfrac{\pi}{2}+k\pi

Those isolated points sit on the $z$ axis, spaced $\pi$ apart, alternating between $f = 1$ and $f = -1$ .

For a concrete evaluation take $(1, 1, 0)$ , where $\sin z = 0$ kills the first two components.

\nabla f(1,1,0) = \left\langle 0,\; 0,\; e \right\rangle \approx \left\langle 0,\; 0,\; 2.7183 \right\rangle

That says something readable about the surface: on the plane $z = 0$ the function is identically zero, so it cannot change in $x$ or $y$ there, and the only way to move off that flat sheet is in the $z$ direction.

The mistake: losing a factor when the variable is not the live one

The dominant errors are dropped factors, and each has a different cause.

Writing $\frac{\partial f}{\partial z} = \cos z$ , having discarded $e^{xy}$ because it contains no $z$ . A factor without the live variable is a multiplier, not something to delete.
Writing $\frac{\partial f}{\partial x} = e^{xy}\sin z$ , forgetting the inner derivative of the exponent. That is the same slip as writing the derivative of $e^{xy}$ as $e^{xy}$ .
Applying the product rule across $e^{xy}$ and $\sin z$ when differentiating in $x$ , and picking up a phantom term $e^{xy}\cos z$ . With respect to $x$ , the derivative of $\sin z$ is zero, so that term vanishes.

Test at $(0,0,0)$ : the true gradient is $\left\langle 0, 0, 1\right\rangle$ , since $\sin 0 = 0$ and $e^{0} = 1$ . The dropped-exponential version also gives $1$ there, so use $(1,1,0)$ instead, where the true $z$ component is $e \approx 2.7183$ and that version returns $1$ .

Frequently asked questions

Why do the $x$ and $y$ components look identical apart from the swap?

Because $f$ is symmetric in $x$ and $y$ : the only place they appear is in the product $xy$ . Swapping them leaves the function unchanged, so it must swap the two partials as well, and $y e^{xy}\sin z$ becomes $x e^{xy}\sin z$ .

Does the gradient ever point purely in the $z$ direction?

Yes, everywhere on the planes $z = k\pi$ , where $\sin z = 0$ kills the first two components while $\cos z = \pm 1$ keeps the third alive. It also happens on the $z$ axis, where $x = y = 0$ makes the first two components vanish, apart from the critical points $z = \frac{\pi}{2}+k\pi$ , where the whole vector is zero and no direction is defined.