Multivariable calculus

Partial Derivatives of e^(x y^2)

Q: Why is there no x in the x partial?

Because u = xy^2 is linear in x, so \frac{\partial u}{\partial x} = y^2 with no x left in it. The variable has not left the problem: it is still sitting inside the factor e^{xy^2}, which is why the x partial still changes value as you move along a horizontal line.

Q: Do the mixed second partials agree?

Yes, and checking that is a free test of both first partials. Differentiating f_x = y^2e^{xy^2} with respect to y gives (2y + 2xy^3)e^{xy^2}, and differentiating f_y = 2xy\,e^{xy^2} with respect to x gives the same expression. If yours disagree, one of the first partials is wrong.

The partials of e^(x y^2) are f_x = y^2 e^(x y^2) and f_y = 2xy e^(x y^2). The exponential reproduces itself, so all the work is in the inner function x y^2. Holding y fixed makes it a constant multiple of x, giving y^2, while holding x fixed leaves x times the derivative of y^2, giving 2xy.

f(x,y) = e^{xy^2}

\frac{\partial f}{\partial x} = y^2\,e^{xy^2}

\frac{\partial f}{\partial y} = 2xy\,e^{xy^2}

The exponential comes along unchanged

Set $u = xy^2$ . Because $\frac{d}{du}e^u = e^u$ , the outer factor of both partials is $e^{xy^2}$ again. Everything that separates the two answers happens inside, in $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ .

\frac{\partial f}{\partial x} = e^{xy^2}\cdot\frac{\partial}{\partial x}(xy^2) = y^2\,e^{xy^2}

Holding $y$ fixed makes $xy^2$ a constant multiple of $x$ , and the constant is $y^2$ . No product rule is needed, because only one factor is moving.

\frac{\partial f}{\partial y} = e^{xy^2}\cdot\frac{\partial}{\partial y}(xy^2) = 2xy\,e^{xy^2}

For $y$ the frozen constant is $x$ , and it multiplies the $2y$ that the power rule produces from $y^2$ . That is where $2xy$ comes from: the $x$ is carried through, not created.

Two ways the inner partial gets truncated

The first error is dropping the frozen variable and writing $\frac{\partial f}{\partial y} = 2y\,e^{xy^2}$ . Test it at $(2,1)$ : the true value is $4e^2 \approx 29.556$ and the truncated one gives $2e^2 \approx 14.778$ , short by exactly the factor $x = 2$ .

The second is swapping the two inner partials. Unlike $e^{xy}$ , this function is not symmetric in $x$ and $y$ , so the swap shows up at almost any test point: at $(2,1)$ the $x$ partial is $1\cdot e^2 \approx 7.389$ while the $y$ partial is about $29.556$ , four times larger.

Name the inner function $u = xy^2$ before differentiating anything.
Freeze one variable and read off exactly what multiplies the other.
Attach the untouched $e^{xy^2}$ last, so it cannot absorb a lost factor.

A whole line of critical points

The exponential is never zero, so $f_x = y^2e^{xy^2}$ vanishes only when $y = 0$ , and at those points $f_y = 2xy\,e^{xy^2}$ is zero too. The critical set is therefore the entire $x$ axis, and along it $f = e^0 = 1$ .

None of those points is a maximum or a minimum. Step off the axis at $x = 1$ and the value rises to $e^{y^2} > 1$ ; do the same at $x = -1$ and it falls to $e^{-y^2} < 1$ . At $(1,\,0.1)$ the value is about $1.0101$ and at $(-1,\,0.1)$ it is about $0.99$ , so the axis is a watershed rather than a ridge or a trough.

One more shape check comes free: $y^2$ is never negative, so $f_x \ge 0$ everywhere and $f$ never decreases as you move right. Any candidate $x$ partial carrying an odd power of $y$ out front would change sign with $y$ and is wrong on that ground alone.

Frequently asked questions

Why is there no x in the x partial?

Because $u = xy^2$ is linear in $x$ , so $\frac{\partial u}{\partial x} = y^2$ with no $x$ left in it. The variable has not left the problem: it is still sitting inside the factor $e^{xy^2}$ , which is why the $x$ partial still changes value as you move along a horizontal line.

Do the mixed second partials agree?

Yes, and checking that is a free test of both first partials. Differentiating $f_x = y^2e^{xy^2}$ with respect to $y$ gives $(2y + 2xy^3)e^{xy^2}$ , and differentiating $f_y = 2xy\,e^{xy^2}$ with respect to $x$ gives the same expression. If yours disagree, one of the first partials is wrong.