Multivariable calculus

Partial Derivatives of e^(xy) cos y

Q: Why does the x-partial have no sine term?

Because \cos y does not depend on x. When you differentiate with respect to x, that factor behaves like a fixed number and comes along unchanged. Sine only appears when you differentiate \cos y, which happens only in the y-partial.

Q: Where does the gradient vanish?

The x-partial y e^{xy}\cos y is zero when y = 0 or \cos y = 0. On the line y = 0 the y-partial reduces to e^{0}(x \cdot 1 - 0) = x, which is zero only at x = 0. On the branch where \cos y = 0 the y-partial is -e^{xy}\sin y, and \sin y = \pm 1 there, so it never vanishes. The origin is the only critical point.

For f(x,y) = e^(xy) cos y, the partial with respect to x is y e^(xy) cos y, and the partial with respect to y is e^(xy) times (x cos y - sin y). Only the y-partial needs the product rule, because cos y is constant when x is the moving variable.

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

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

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

Decide which factors are alive

This is the cleanest example of a rule that saves time: before differentiating, list which factors actually contain the variable you are moving. In $x$ , only $e^{xy}$ contains $x$ , so $\cos y$ is a constant multiplier and the product rule is not needed.

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

In $y$ , both factors are alive: $e^{xy}$ depends on $y$ through the exponent and $\cos y$ obviously does. Now the product rule is unavoidable.

\frac{\partial f}{\partial y} = x e^{xy}\cos y + e^{xy}(-\sin y)

Factor the common $e^{xy}$ out. The exponential is never zero, so the factored form immediately tells you the $y$ -partial vanishes exactly when $x\cos y = \sin y$ , that is when $\tan y = x$ .

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

The mistake: sign and chain rule slips in the y-partial

Two errors dominate here. The first is writing $+\sin y$ instead of $-\sin y$ , because the derivative of cosine carries the minus sign. The second is writing $e^{xy}$ for the derivative of $e^{xy}$ in $y$ , forgetting that the chain rule brings down the inner derivative $x$ .

$\frac{\partial}{\partial y}e^{xy} = x e^{xy}$ , not $e^{xy}$ and not $y e^{xy}$ .
$\frac{\partial}{\partial x}e^{xy} = y e^{xy}$ . The coefficient is always the other variable.
$\frac{d}{dy}\cos y = -\sin y$ . The minus sign is what makes the two terms compete rather than reinforce.

A structural check: the $x$ -partial must contain $\cos y$ untouched, because a constant factor passes straight through differentiation. If your $x$ -partial has a $\sin y$ in it, you differentiated a factor that was supposed to be frozen.

A point where the two partials disagree

At $(1,0)$ the exponential is $e^{0}=1$ , $\cos 0 = 1$ , and $\sin 0 = 0$ . The $x$ -partial picks up the factor $y = 0$ and dies; the $y$ -partial survives.

\frac{\partial f}{\partial x}(1,0) = 0, \qquad \frac{\partial f}{\partial y}(1,0) = 1

This is a good reminder that a partial derivative can be zero without anything special happening to the function. Along the line $y = 0$ the function is constantly 1, so of course moving in $x$ changes nothing there.

Frequently asked questions

Why does the x-partial have no sine term?

Because $\cos y$ does not depend on $x$ . When you differentiate with respect to $x$ , that factor behaves like a fixed number and comes along unchanged. Sine only appears when you differentiate $\cos y$ , which happens only in the $y$ -partial.

Where does the gradient vanish?

The $x$ -partial $y e^{xy}\cos y$ is zero when $y = 0$ or $\cos y = 0$ . On the line $y = 0$ the $y$ -partial reduces to $e^{0}(x \cdot 1 - 0) = x$ , which is zero only at $x = 0$ . On the branch where $\cos y = 0$ the $y$ -partial is $-e^{xy}\sin y$ , and $\sin y = \pm 1$ there, so it never vanishes. The origin is the only critical point.