A first-order differential equation is frequently written \dfrac{dy}{dx}=f(x,y) — a right-hand side depending on two variables, even though only one of them, x, is being differentiated with respect to. Reading that right side correctly, and later checking whether it's exact by comparing two partial derivatives, both require the single skill this section refreshes.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
For f(x,y) = \dfrac{x}{x+y}, find f_x(1,1).
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Why
Quotient rule with y fixed: f_x = \dfrac{(x+y)(1) - x(1)}{(x+y)^2} = \dfrac{y}{(x+y)^2}, so f_x(1,1) = \dfrac{1}{4} = 0.25.
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Why
Use the quotient rule treating y as a constant.
f_x = y / (x+y)^2.
From Calculus III — Multivariable Calculus
For f(x,y) = \sin(xy), find f_x(x,y).
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Chain rule with y frozen as a constant: derivative of \sin(u) is \cos(u)\cdot u', where u=xy and u_x = y.
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Why
Chain rule: d/dx sin(xy) = cos(xy) * d/dx(xy).
From Calculus III — Multivariable Calculus
For f(x,y,z) = x^2yz^3, find f_z(x,y,z).
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Freeze x and y; the power rule on z^3 gives 3z^2.
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Why
This works the same way with three variables -- freeze all but the one you're differentiating with respect to.
Freezing every variable except one turns a two-variable problem into an ordinary single-variable derivative you already know how to take. A partial derivative is nothing more than that freeze, applied once per variable.
The partial derivative \dfrac{\partial f}{\partial x} treats every other variable as a fixed constant and differentiates as usual in x alone; \dfrac{\partial f}{\partial y} does the same, freezing x instead. For f(x,y)=x^2y+\sin(xy), f_x = 2xy+y\cos(xy) (the \sin term differentiated by the chain rule, with y frozen as a constant coefficient) and f_y = x^2+x\cos(xy). This is exactly the computation behind checking exactness later in this course: an equation M(x,y)\,dx+N(x,y)\,dy=0 is exact precisely when \dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}, two partial derivatives compared against each other.
Two frozen slices, two partial derivatives
Watch the tangent plane at the marked point -- its tilt along the x-direction is f_x, its tilt along the y-direction is f_y.
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Worked example
Both partial derivatives of one function
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1
For f(x,y)=x^2y+y^3, find f_x and f_y.
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2
For f_x, treat y as a constant: f_x = 2xy. For f_y, treat x as a constant: f_y = x^2+3y^2.
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3\boxed{f_x=2xy,\qquad f_y=x^2+3y^2}
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Worked example
Reading dy/dx = f(x,y) as a slope that depends on both coordinates
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1
For \dfrac{dy}{dx}=x-y, what is the slope of a solution curve at the point (2,3)?
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2
Substitute the point directly into the right-hand side — no differentiation is needed here, only evaluation.
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3
\dfrac{dy}{dx}\Big|_{(2,3)} = 2-3=-1. \boxed{-1}
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Where this goes next
Exact equations, ahead
Chapter 2's exactness test is a straight partial-derivative comparison: compute \partial M/\partial y and \partial N/\partial x from the two sides of the differential form, and check whether they match. That entire method rests on the single skill in this section.
Check your understanding (counts toward your progress)
For g(x,y)=x^3-4xy^2, find g_y.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Treating x as a constant, \dfrac{\partial}{\partial y}[x^3]=0 and \dfrac{\partial}{\partial y}[-4xy^2]=-8xy.
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Why
For \dfrac{dy}{dx}=x^2-y, what is the slope of a solution curve at the point (3,4)?
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Why
Substitute directly: x^2-y=9-4=5.
Correct answer
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Why
Substitute x=3,y=4 directly into x^2-y -- no differentiation is needed here.
To find \partial M/\partial y for M(x,y)=x^2y+\sin x, which variable is treated as a fixed constant?
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y is the variable being differentiated with respect to -- it can't also be frozen.
Exactly one variable is frozen at a time; the other is the one you differentiate.
A partial derivative always freezes every variable except the one in the denominator.
Why
\partial/\partial y differentiates with respect to y while treating x as a fixed constant -- giving x^2.
The variable being differentiated with respect to can't also be the one held fixed.
For f(x,y)=3x^2y-y^3, find f_x.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Treating y as a constant, \dfrac{\partial}{\partial x}[3x^2y]=6xy and \dfrac{\partial}{\partial x}[-y^3]=0.
Correct answer
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Why