Multivariable calculus

Partial Derivatives of x^2 y e^x

Q: Where is the x-partial equal to zero?

Set xy(x+2)e^{x} = 0. Since e^{x} is never zero, the zeros are x = 0, x = -2, and y = 0. Combining that with \frac{\partial f}{\partial y} = x^{2}e^{x} = 0, which forces x = 0, the critical points are the whole line x = 0.

For f(x,y) = x^2 y e^x, the partial with respect to x is x y (x + 2) e^x, and the partial with respect to y is x^2 e^x. Differentiating in x needs the product rule on x^2 times e^x with y held fixed. Differentiating in y does not, because f is linear in y.

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

\frac{\partial f}{\partial x} = xy(x+2)e^{x}

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

Only one of the two partials needs the product rule

Before you reach for a rule, ask which variable is moving. To find $\frac{\partial f}{\partial x}$ , freeze $y$ . What is left is the constant $y$ multiplying $x^{2}e^{x}$ , and that is a genuine product of two functions of $x$ , so the product rule applies.

\frac{\partial f}{\partial x} = y\left(2x\,e^{x} + x^{2}e^{x}\right)

Both terms share $x e^{x}$ , so factor it out. Factoring is worth the extra line: the factored form is what you need later when you set the gradient to zero.

\frac{\partial f}{\partial x} = xy(x+2)e^{x}

Now freeze $x$ instead. With $x$ constant, $x^{2}e^{x}$ is one fixed number and $f$ is that number times $y$ , a straight line in $y$ . The derivative of a constant times $y$ is the constant, so no product rule is involved at all.

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

The mistake: differentiating every factor at once

The usual error is to see three factors, $x^{2}$ , $y$ , and $e^{x}$ , and try to apply a three-way product rule in $x$ . That produces a stray term where $y$ has been differentiated to 1, and the answer picks up an extra $x^{2}e^{x}$ that does not belong.

Hold the line on this: a partial derivative in $x$ treats $y$ exactly like the number 7. You would never differentiate the 7 in $7x^{2}e^{x}$ , and $y$ gets the same treatment.

In $\frac{\partial f}{\partial x}$ , the product rule pairs $x^{2}$ with $e^{x}$ only. The $y$ rides along as a constant multiplier.
In $\frac{\partial f}{\partial y}$ , there is no product of two $y$ -dependent factors, so the product rule has nothing to act on.
A quick sanity check: $\frac{\partial f}{\partial y}$ must not contain $y$ , because $f$ is degree one in $y$ .

Check the answer at a point

Evaluating at $(1,1)$ gives numbers you can hold in your head. From the factored form, $\frac{\partial f}{\partial x}(1,1) = (1)(1)(3)e = 3e$ , and $\frac{\partial f}{\partial y}(1,1) = e$ .

\nabla f(1,1) = \langle 3e,\; e \rangle

The ratio 3 to 1 says the surface climbs three times as fast in the $x$ direction as in the $y$ direction at that point, which matches the fact that $x$ appears both as a power and inside the exponential while $y$ appears only once.

Frequently asked questions

Why does the x-partial need the product rule but the y-partial does not?

Because the product rule only fires when two factors both depend on the variable you are differentiating. In $x$ , both $x^{2}$ and $e^{x}$ depend on $x$ . In $y$ , only the single factor $y$ depends on $y$ , and $x^{2}e^{x}$ is a frozen constant, so the derivative is that constant.

Where is the x-partial equal to zero?

Set $xy(x+2)e^{x} = 0$ . Since $e^{x}$ is never zero, the zeros are $x = 0$ , $x = -2$ , and $y = 0$ . Combining that with $\frac{\partial f}{\partial y} = x^{2}e^{x} = 0$ , which forces $x = 0$ , the critical points are the whole line $x = 0$ .