Multivariable calculus

Gradient of e^x cos(y): Components and Magnitude

Q: Why does the magnitude of the gradient not depend on y?

Because the two components are e^{x}\cos y and -e^{x}\sin y, and squaring and adding them gives e^{2x}(\cos^{2}y + \sin^{2}y) = e^{2x}. The y dependence sits entirely in the direction of the vector, not its length.

Q: Does e^{x}\cos y have any maximum or minimum?

No. A local extremum needs \nabla f = \left\langle 0,0\right\rangle, and that would force \cos y = 0 and \sin y = 0 at the same y, which is impossible. On any closed bounded region the extreme values are therefore on the boundary.

For f(x, y) = e^x cos(y) the gradient is the vector (e^x cos y, -e^x sin y). Each partial treats the other variable as a constant, so the x partial reproduces the function and the y partial picks up the minus sign from the derivative of cosine. The length of this gradient is e^x at every point.

f(x,y) = e^{x}\cos y

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

\frac{\partial f}{\partial x} = e^{x}\cos y

\frac{\partial f}{\partial y} = -e^{x}\sin y

A product where each factor owns one variable

No chain rule and no product rule are needed here. The factor $\cos y$ contains no $x$ , so for $\frac{\partial f}{\partial x}$ it is just a constant multiplier sitting in front of $e^{x}$ .

\frac{\partial f}{\partial x} = \cos y \cdot \frac{d}{dx}\left(e^{x}\right) = e^{x}\cos y

Swap roles for the other component. Now $e^{x}$ is the constant and $\cos y$ is live, and the derivative of cosine carries a minus sign.

\frac{\partial f}{\partial y} = e^{x}\cdot \frac{d}{dy}\left(\cos y\right) = -e^{x}\sin y

The $x$ partial coming back as $f$ itself is not a coincidence to be suspicious of. Differentiating $e^{x}$ returns $e^{x}$ , and the $y$ factor never moved.

The length is e^x, and the direction spins with y

Factor $e^{x}$ out of the gradient and the Pythagorean identity finishes the magnitude.

\left|\nabla f\right| = e^{x}\sqrt{\cos^{2} y + \sin^{2} y} = e^{x}

So the steepness of this surface depends only on $x$ . Move along a vertical line and the surface rises and falls, yet the maximum rate of change never changes size; only the direction of steepest ascent rotates, tracing the unit vector $\left\langle \cos y, -\sin y\right\rangle$ as $y$ increases.

The gradient is never the zero vector, since $e^{x} > 0$ and $\cos y$ and $\sin y$ never vanish at the same $y$ . This surface has no critical points at all.

One more property worth carrying forward: the second partials cancel, $f_{xx} + f_{yy} = e^{x}\cos y - e^{x}\cos y = 0$ , so $f$ is harmonic. It is the real part of $e^{z}$ in complex variables, which is where this example usually reappears.

The mistake: losing the minus sign, or chain-ruling the exponential

Two errors dominate on this page.

Writing $\frac{\partial f}{\partial y} = e^{x}\sin y$ without the minus. Then the gradient points the wrong way in the $y$ direction, and the linear approximation predicts the surface rising where it actually falls.
Writing $\frac{\partial f}{\partial y} = e^{x}\cos y \cdot (\text{something})$ , as if the exponential needed a chain rule in $y$ . Its exponent is $x$ alone, so with respect to $y$ it is a frozen constant.

Check at $(0, 0)$ : $\nabla f(0,0) = \left\langle 1, 0 \right\rangle$ . The surface at that point is level in the $y$ direction, which matches $\cos y$ having a maximum at $y = 0$ . A wrong sign would not show up here, so test at $(0, \frac{\pi}{2})$ as well, where $\nabla f = \left\langle 0, -1 \right\rangle$ and $f$ is clearly decreasing in $y$ .

Frequently asked questions

Why does the magnitude of the gradient not depend on $y$ ?

Because the two components are $e^{x}\cos y$ and $-e^{x}\sin y$ , and squaring and adding them gives $e^{2x}(\cos^{2}y + \sin^{2}y) = e^{2x}$ . The $y$ dependence sits entirely in the direction of the vector, not its length.

Does $e^{x}\cos y$ have any maximum or minimum?

No. A local extremum needs $\nabla f = \left\langle 0,0\right\rangle$ , and that would force $\cos y = 0$ and $\sin y = 0$ at the same $y$ , which is impossible. On any closed bounded region the extreme values are therefore on the boundary.