Multivariable calculus

Gradient of e^x sin(y) cos(z): Separable Product

Q: Why does \frac{\partial f}{\partial x} equal f itself?

Because the only x dependence is the factor e^x, which is its own derivative. Differentiating it reproduces the same product, so this function satisfies \frac{\partial f}{\partial x} = f at every point.

Q: Where is the gradient zero?

Where \sin y = 0 and \cos z = 0 at the same time, that is y = k\pi and z = \frac{\pi}{2} + m\pi. Since e^x is never zero and \sin y and \cos y never vanish together, the middle component forces \cos z = 0, and then the third component forces \sin y = 0. The critical set is a grid of lines parallel to the x-axis, and f = 0 on all of them.

For f(x, y, z) = e^x sin(y) cos(z) the gradient is (e^x sin(y) cos(z), e^x cos(y) cos(z), -e^x sin(y) sin(z)). Each component differentiates exactly one factor and copies the other two. The x component equals the function itself. At (0, pi/2, 0) the gradient is (1, 0, 0).

f(x,y,z) = e^x \sin y \cos z

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

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

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

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

A separable product: change one factor, copy the rest

This function is a product of three pieces, each depending on a single variable. That structure makes the gradient easy: in each component, differentiate the factor that owns that variable and leave the other two exactly as they are.

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

Since $\frac{d}{dx}e^x = e^x$ , the $x$ component is the original function again. That is a useful check: for this $f$ , $\frac{\partial f}{\partial x} = f$ everywhere.

\frac{\partial f}{\partial y} = e^x \cos y \cos z, \qquad \frac{\partial f}{\partial z} = -e^x \sin y \sin z

Only the $z$ component picks up a sign, because $\frac{d}{dz}\cos z = -\sin z$ .

The mistake: the lost minus sign, and the phantom product rule

Two errors dominate here. The first is writing $\frac{\partial f}{\partial z} = e^x \sin y \sin z$ without the minus. That single sign flips the direction of steepest ascent, so an otherwise perfect answer points the wrong way.

The second is applying the product rule across all three factors, producing a sum of three terms in every component. The product rule is only needed when more than one factor depends on the variable you are differentiating. Here exactly one does, so the other two are constants and their derivatives are zero.

In $\frac{\partial f}{\partial y}$ , both $e^x$ and $\cos z$ are constants, so no extra terms appear.
Sign check: $\frac{d}{dy}\sin y = \cos y$ is positive in the formula, $\frac{d}{dz}\cos z = -\sin z$ is negative.
If a component of your gradient has three added terms, you have used the product rule where it was not needed.

Reading the gradient at a point

Take $(0, \frac{\pi}{2}, 0)$ , where $e^0 = 1$ , $\sin\frac{\pi}{2} = 1$ and $\cos 0 = 1$ , so $f = 1$ . This is a maximum of the trigonometric part.

\nabla f\left(0, \tfrac{\pi}{2}, 0\right) = \left\langle 1,\; 0,\; 0 \right\rangle

The $y$ and $z$ components are zero because $\cos\frac{\pi}{2} = 0$ and $\sin 0 = 0$ : the trigonometric factors are each sitting at a peak, so a small move in $y$ or $z$ barely changes $f$ . Only growth in $x$ still pays, at rate 1.

The magnitude has a tidy closed form. Squaring and adding, the $\sin^2 y + \cos^2 y$ pair collapses and you are left with a two term expression.

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

Frequently asked questions

Why does $\frac{\partial f}{\partial x}$ equal $f$ itself?

Because the only $x$ dependence is the factor $e^x$ , which is its own derivative. Differentiating it reproduces the same product, so this function satisfies $\frac{\partial f}{\partial x} = f$ at every point.

Where is the gradient zero?

Where $\sin y = 0$ and $\cos z = 0$ at the same time, that is $y = k\pi$ and $z = \frac{\pi}{2} + m\pi$ . Since $e^x$ is never zero and $\sin y$ and $\cos y$ never vanish together, the middle component forces $\cos z = 0$ , and then the third component forces $\sin y = 0$ . The critical set is a grid of lines parallel to the $x$ -axis, and $f = 0$ on all of them.