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Partials, explained

The reasoning behind practice exam Q11 (contour maps) and Q12 (chain rule with a table), plus the rest of the partial-derivative toolkit.

Start here

What a partial derivative actually is

f_x = how fast the height changes when you walk in the +x direction only, with y frozen. f_y is the same thing walking in +y.

Every partial is just rise ÷ run along one direction.

Whether you're given a formula, a table, a contour map, or a 3D graph, you're computing that same rise-over-run. Only the way you read "rise" changes.

Exam Q11

Partials from a contour map

On a contour map, the labels are heights. So:

∂g/∂x ≈ (change in label) ÷ (distance walked right)
∂g/∂y ≈ (change in label) ÷ (distance walked up)

Recipe:

Find the point and read its label. At (2, 2) it's 6.
For g_x: slide along the horizontal line through the point and write down the label at each gridline.
```
y = 2:   x=1 → 7    x=2 → 6    x=3 → 5
g_x = (5 − 7)/(3 − 1) = −1
```

For g_y: slide along the vertical line.

x = 2:   y=1.5 → 5    y=2 → 6    y=2.5 → 7
g_y = (7 − 5)/(2.5 − 1.5) = 2

Sign shortcut: labels go up in the direction you walk → positive. They go down → negative. Curves packed tight → steep, so bigger size. Spread out → gentle.

Trap: divide by the distance, not the number of curves. Two labels across ½ unit is a slope of 2, not 1.

Exam Q11(c)

The mixed partial on a contour map

∂²g/∂y∂x = ∂/∂y (g_x). Read it as: as I move UP, does the x-slope get bigger or smaller?

Measure g_x at a row below the point and a row above it, the same way as before.
Compare them.

y = 1:  curves tight horizontally    g_x ≈ −1.5
y = 2:                               g_x ≈ −1
y = 3:  9 at x=0, 8 at x=2 (spread)  g_x ≈ −0.5

going up, g_x goes −1.5 → −1 → −0.5  (increasing)
⇒ ∂²g/∂y∂x is POSITIVE

What you see: horizontal spacing widening as you go up means the slope is getting gentler. For negative slopes, gentler means less negative, which means increasing. So the answer is positive.

Trap: "less steep" doesn't mean "decreasing". −0.5 is bigger than −1.5. Track the actual signed numbers.

Exam Q12

Chain rule with a table

Setup: F(s,t) = f(x(s,t), y(s,t)). F depends on t in two ways: through x and through y.

          F
        /   \
       x     y
      / \   / \
     s   t s   t

∂F/∂t = f_x · x_t  +  f_y · y_t
        (path via x)  (path via y)

Multiply along each path from F down to t, then add the paths.

Worked (x = s + 2t, y = s² − t, at (s,t) = (2,1)):

Find the (x, y) point first.

x = 2 + 2(1) = 4
y = 2² − 1    = 3     → use the (4, 3) row

Inner derivatives (from the x and y formulas):
```
x_t = 2        y_t = −1
```
Outer derivatives (from the table, row (4, 3)):
```
f_x = 3        f_y = 1
```
Combine:
```
∂F/∂t = 3·2 + 1·(−1) = 5
```

THE TRAP: the table has a (2, 1) row sitting right there. It's a decoy. f only knows about x and y, so its partials are always read at the (x, y) point you calculate, never at (s, t). The inner derivatives x_t, y_t are the ones that use s and t.

Which rows are decoys? Usually one row is the (s,t) point, and one is the (x,y) point with x and y swapped. Compute (x, y) first and you can't fall for either.

Exam Q10

Signs from a 3D graph

f_x sign: stand at P and walk in +x (follow the x-axis arrow). Uphill → +, downhill → −, level (on a crest or in a trough) → 0.
f_y sign: same, walking in +y.
f_xx / f_yy sign: the concavity of that same walk. Bowl-shaped (valley) → +, dome-shaped (hilltop) → −.

First derivative = which way you're going. Second derivative = which way it bends.

Trap: the answer choices include 0 on purpose. If P sits on the very top of a ridge, the slope across the ridge is 0 and the concavity is negative.

From a table (HW8 wind-chill)

Estimating a partial from a table

Stay in the row or column your point is in. The variable you're NOT differentiating stays fixed.
Take the two immediate neighbors on either side (centered difference).
Rise ÷ run.

f_T(−15, 30):  stay in the v = 30 column
  T = −10 → −20     T = −20 → −33
  (−20 − (−33)) / (−10 − (−20)) = 13/10 = 1.30

Trap: subtracting negatives. −20 − (−33) = +13.

Formula

Mixed partials from a formula + the limit definition

f = x²y³
f_x  = 2xy³         (y is a constant)
f_xy = 6xy²         (now differentiate f_x by y)
f_yx = 6xy²         (Clairaut: same answer)

When the formula gives 0/0 at a point, go back to the definition:

f_x(a,b) = lim  [ f(a+h, b) − f(a, b) ] / h
           h→0

f = ∛(x³+y³):  f(h,0) = h,  f(0,0) = 0
f_x(0,0) = lim h/h = 1

MATH 241 · Midterm 1 · Stewart 14.3–14.5

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Formula sheet

Partials, explained

What a partial derivative actually is📋 Copy for AI

Partials from a contour map📋 Copy for AI

The mixed partial on a contour map📋 Copy for AI

Chain rule with a table📋 Copy for AI

Signs from a 3D graph📋 Copy for AI

Estimating a partial from a table📋 Copy for AI