Multivariable calculus

Gradient of x e^(yz): Chain Rule in the Exponent

Q: Why does the x partial have no exponential factor multiplying it?

It does have one: \frac{\partial f}{\partial x} = e^{yz} is entirely that factor. Since f is linear in x, differentiating removes the x and leaves the coefficient, which is e^{yz}.

Q: What is the direction of steepest increase at (1,1,1)?

There e^{yz} = e, so \nabla f = \left\langle e, e, e \right\rangle, which points along \left\langle 1,1,1 \right\rangle. The steepest ascent direction is the unit vector \frac{1}{\sqrt{3}}\left\langle 1,1,1 \right\rangle and the maximum rate is e\sqrt{3} \approx 4.71.

For f(x, y, z) = x e^(yz) the gradient is (e^(yz), xz e^(yz), xy e^(yz)). The x component is simple because the exponential is a constant multiplier there, while the y and z components each pick up an inner factor from differentiating the exponent yz. At (2, 1, 0) the gradient is (1, 0, 2).

f(x,y,z) = x e^{yz}

\nabla f = \left\langle e^{yz},\; xz e^{yz},\; xy e^{yz} \right\rangle

\frac{\partial f}{\partial x} = e^{yz}

\frac{\partial f}{\partial y} = xz e^{yz}

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

Three different jobs for three variables

Each variable plays a different role, so each partial uses a different rule. With $y$ and $z$ frozen, $e^{yz}$ is one fixed number multiplying $x$ , and differentiating a constant times $x$ returns the constant.

\frac{\partial f}{\partial x} = e^{yz}

For $y$ , the outer function is the exponential and the inner function is the exponent $yz$ . The chain rule says to copy the exponential and multiply by the derivative of the exponent, which is $z$ when $z$ is held fixed.

\frac{\partial f}{\partial y} = x \cdot e^{yz} \cdot \frac{\partial}{\partial y}\left(yz\right) = xz e^{yz}

The $z$ partial is the mirror image, with inner derivative $y$ .

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

The mistake: dropping the inner factor

The classic wrong answer is $\frac{\partial f}{\partial y} = x e^{yz}$ . It looks reasonable because $e^u$ differentiates to $e^u$ , but that rule applies only when $u$ is the variable itself. Here $u = yz$ , and the chain rule demands the extra factor $\frac{\partial u}{\partial y} = z$ .

Structure check: the exponent $yz$ changes at rate $z$ when $y$ moves, so a $z$ has to appear in the $y$ partial. At $z = 0$ the exponent does not move at all as $y$ changes, and the $y$ partial is correctly zero there.
The forgotten factor is invisible when $z = 1$ , which is why testing at $z = 1$ hides the error. Test at $z = 2$ instead.
A second slip is copying the same inner factor into both components. The $y$ partial gets $z$ , the $z$ partial gets $y$ , never the same one twice.

If you are unsure, rename the exponent. Write $u = yz$ and $f = xe^u$ , then $\frac{\partial f}{\partial y} = xe^u \frac{\partial u}{\partial y}$ . The structure becomes hard to misread.

A check, and why this surface has no critical points

At $(2, 1, 0)$ the exponent is $yz = 0$ , so $e^{yz} = 1$ and $f = 2$ .

\nabla f(2,1,0) = \left\langle 1,\; 2 \cdot 0 \cdot 1,\; 2 \cdot 1 \cdot 1 \right\rangle = \left\langle 1,\; 0,\; 2 \right\rangle

The $y$ component is zero here because $z = 0$ : with $z$ pinned at zero, changing $y$ does not change the exponent, so $f$ is momentarily flat in that direction. This is a genuine feature of the function, not an arithmetic accident.

The $x$ component is $e^{yz}$ , and an exponential is never zero. So $\nabla f$ is never the zero vector and $f$ has no critical points anywhere in space. Any maximum or minimum of $f$ on a bounded region must therefore occur on the boundary.

Frequently asked questions

Why does the $x$ partial have no exponential factor multiplying it?

It does have one: $\frac{\partial f}{\partial x} = e^{yz}$ is entirely that factor. Since $f$ is linear in $x$ , differentiating removes the $x$ and leaves the coefficient, which is $e^{yz}$ .

What is the direction of steepest increase at $(1,1,1)$ ?

There $e^{yz} = e$ , so $\nabla f = \left\langle e, e, e \right\rangle$ , which points along $\left\langle 1,1,1 \right\rangle$ . The steepest ascent direction is the unit vector $\frac{1}{\sqrt{3}}\left\langle 1,1,1 \right\rangle$ and the maximum rate is $e\sqrt{3} \approx 4.71$ .