Multivariable calculus

Partial Derivatives of sin(x)/(1 + y^2)

Q: Why does the y-partial have a squared denominator but the x-partial does not?

The quotient rule always produces the square, but in the x-partial the surviving term still has a full factor of 1+y^{2} in the numerator, which cancels one copy. In the y-partial nothing cancels, because the numerator term that would cancel is the one that vanished.

Q: Where is this function largest?

The numerator is largest when \sin x = 1 and the denominator is smallest when y = 0, so the maximum value 1 occurs at points such as \left(\frac{\pi}{2}, 0\right). Both partials are zero there, which is the algebraic confirmation.

For f(x,y) = sin(x)/(1 + y^2), the partial with respect to x is cos(x)/(1 + y^2), and the partial with respect to y is -2y sin(x)/(1 + y^2) squared. Only the y-partial needs the quotient rule, since in x the denominator is a frozen constant.

f(x,y) = \frac{\sin x}{1 + y^{2}}

\frac{\partial f}{\partial x} = \frac{\cos x}{1 + y^{2}}

\frac{\partial f}{\partial y} = \frac{-2y\sin x}{\left(1+y^{2}\right)^{2}}

A separated function makes each partial easy

This function factors as a function of $x$ times a function of $y$ , namely $\sin x$ times $\left(1+y^{2}\right)^{-1}$ . Whenever that happens, each partial keeps one factor untouched and differentiates the other.

\frac{\partial f}{\partial x} = \frac{1}{1+y^{2}} \cdot \cos x = \frac{\cos x}{1+y^{2}}

For $\frac{\partial f}{\partial y}$ the numerator $\sin x$ is constant, so the quotient rule collapses: the derivative of the top is zero and only the second term survives.

\frac{\partial f}{\partial y} = \frac{0 \cdot \left(1+y^{2}\right) - \sin x \cdot 2y}{\left(1+y^{2}\right)^{2}}

\frac{\partial f}{\partial y} = \frac{-2y\sin x}{\left(1+y^{2}\right)^{2}}

You could also write $f = \sin x \cdot \left(1+y^{2}\right)^{-1}$ and use the power and chain rules, which gives $\sin x \cdot (-1)\left(1+y^{2}\right)^{-2}(2y)$ , the same thing.

The mistake: using the quotient rule where a constant multiple would do

Running the full quotient rule for the $x$ -partial is not wrong, but it is where sign errors get born. The denominator has $x$ -derivative zero, so the subtracted term disappears and one factor of $1+y^{2}$ cancels. Skipping straight to a constant multiple is faster and safer.

In the $x$ -partial the denominator is squared only if you go through the quotient rule and forget to cancel. The simplified answer has $1+y^{2}$ to the first power.
In the $y$ -partial the minus sign comes from differentiating the denominator, not from the sine. Sine is untouched there.
Since $1+y^{2} \ge 1$ , nothing is ever undefined and no domain restrictions are needed.

A quick sanity check on the $x$ -partial: at any $x$ where $\cos x = 0$ , the whole $x$ -partial must be zero regardless of $y$ . If your expression still depends on $y$ in a way that survives there, something went wrong.

Reading the gradient at a point

At $(0,1)$ we have $\sin 0 = 0$ and $\cos 0 = 1$ , with $1+y^{2} = 2$ .

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

The $y$ -component is zero on the whole line $x = 0$ , because every term in the $y$ -partial carries the factor $\sin x$ . On that line the surface is flat in the $y$ direction, sitting at height zero.

Frequently asked questions

Why does the y-partial have a squared denominator but the x-partial does not?

The quotient rule always produces the square, but in the $x$ -partial the surviving term still has a full factor of $1+y^{2}$ in the numerator, which cancels one copy. In the $y$ -partial nothing cancels, because the numerator term that would cancel is the one that vanished.

Where is this function largest?

The numerator is largest when $\sin x = 1$ and the denominator is smallest when $y = 0$ , so the maximum value 1 occurs at points such as $\left(\frac{\pi}{2}, 0\right)$ . Both partials are zero there, which is the algebraic confirmation.