Chain Rule for Partial Derivatives: Changing Variables (Polar Coordinates)
I Key FormulasKey Insight: The result is a sum of products, one per intermediate variable, and the curly ∂ \partial ∂ ∂ f ∂ u = ∂ f ∂ u + ∂ f ∂ u \frac{\partial f}{\partial u} = \frac{\partial f}{\partial u} + \frac{\partial f}{\partial u} ∂ u ∂ f = ∂ u ∂ f + ∂ u ∂ f ∂ f ∂ u = 0 \frac{\partial f}{\partial u} = 0 ∂ u ∂ f = 0
II The Polar Rates at a PointIncreasing r r r P P P ( cos θ , sin θ ) (\cos\theta, \sin\theta) ( cos θ , sin θ ) θ \theta θ ( − r sin θ , r cos θ ) (-r\sin\theta,\ r\cos\theta) ( − r sin θ , r cos θ )
III One Product per PathFor f = x y f = xy f = x y f f f
IV The Rate as a LimitThe chain-rule value is the limit of Δ f / Δ r \Delta f/\Delta r Δ f /Δ r Δ f / Δ θ \Delta f/\Delta\theta Δ f /Δ θ
Move
along the ray (Δr)
around the circle (Δθ)
Step
▶ play
V Quiz Questions
Problem 1 · The Radial Rate
Given: f ( x , y ) = x 2 + y 2 f(x, y) = x^2 + y^2 f ( x , y ) = x 2 + y 2 x = r cos θ x = r\cos\theta x = r cos θ y = r sin θ y = r\sin\theta y = r sin θ find f r f_r f r f r = cos θ f x + sin θ f y f_r = \cos\theta\, f_x + \sin\theta\, f_y f r = cos θ f x + sin θ f y
a f r = 2 r cos θ + 2 r sin θ f_r = 2r\cos\theta + 2r\sin\theta f r = 2 r cos θ + 2 r sin θ b f r = 2 r f_r = 2r f r = 2 r c f r = 0 f_r = 0 f r = 0 d f r = 2 r cos 2 θ f_r = 2r\cos^2\theta f r = 2 r cos 2 θ
✅ Correct! f r = cos θ ( 2 r cos θ ) + sin θ ( 2 r sin θ ) = 2 r f_r = \cos\theta(2r\cos\theta) + \sin\theta(2r\sin\theta) = 2r f r = cos θ ( 2 r cos θ ) + sin θ ( 2 r sin θ ) = 2 r , matching
f = r 2 f = r^2 f = r 2 directly.
❌ The rates are missing. That is
f x + f y f_x + f_y f x + f y ; each partial must be multiplied by its rate,
x r = cos θ x_r = \cos\theta x r = cos θ and
y r = sin θ y_r = \sin\theta y r = sin θ .
❌ That is f θ f_\theta f θ The function is constant around each circle, so
f θ = 0 f_\theta = 0 f θ = 0 , but moving out along the ray increases it.
❌ Only one path. 2 r cos 2 θ 2r\cos^2\theta 2 r cos 2 θ is the path through
x x x alone; increasing
r r r moves
y y y as well, adding
sin θ ⋅ 2 r sin θ \sin\theta \cdot 2r\sin\theta sin θ ⋅ 2 r sin θ .
Show solution
The partials are f x = 2 x = 2 r cos θ f_x = 2x = 2r\cos\theta f x = 2 x = 2 r cos θ f y = 2 y = 2 r sin θ f_y = 2y = 2r\sin\theta f y = 2 y = 2 r sin θ
f r = cos θ ( 2 r cos θ ) + sin θ ( 2 r sin θ ) = 2 r ( cos 2 θ + sin 2 θ ) = 2 r . f_r = \cos\theta\,(2r\cos\theta) + \sin\theta\,(2r\sin\theta) = 2r(\cos^2\theta + \sin^2\theta) = 2r. f r = cos θ ( 2 r cos θ ) + sin θ ( 2 r sin θ ) = 2 r ( cos 2 θ + sin 2 θ ) = 2 r . Check by substitution: f = r 2 cos 2 θ + r 2 sin 2 θ = r 2 f = r^2\cos^2\theta + r^2\sin^2\theta = r^2 f = r 2 cos 2 θ + r 2 sin 2 θ = r 2 f r = 2 r f_r = 2r f r = 2 r
Problem 2 · No Cancelling
Given: f ( x , y ) = x y f(x, y) = xy f ( x , y ) = x y x = u + v x = u + v x = u + v y = u − v y = u - v y = u − v find ∂ f ∂ u \dfrac{\partial f}{\partial u} ∂ u ∂ f ( u , v ) = ( 2 , 1 ) (u, v) = (2, 1) ( u , v ) = ( 2 , 1 )
a ∂ f ∂ u = 3 \dfrac{\partial f}{\partial u} = 3 ∂ u ∂ f = 3 b ∂ f ∂ u = 8 \dfrac{\partial f}{\partial u} = 8 ∂ u ∂ f = 8 c ∂ f ∂ u = 4 \dfrac{\partial f}{\partial u} = 4 ∂ u ∂ f = 4 d ∂ f ∂ u = − 2 \dfrac{\partial f}{\partial u} = -2 ∂ u ∂ f = − 2
✅ Correct! Both paths count:
f x x u + f y y u = ( 1 ) ( 1 ) + ( 3 ) ( 1 ) = 4 f_x x_u + f_y y_u = (1)(1) + (3)(1) = 4 f x x u + f y y u = ( 1 ) ( 1 ) + ( 3 ) ( 1 ) = 4 .
❌ Only one path. 3 3 3 is the path through
y y y alone; changing
u u u moves
x x x too, and the path through
x x x adds
( 1 ) ( 1 ) (1)(1) ( 1 ) ( 1 ) .
❌ That is the cancellation. Cancelling
∂ x \partial x ∂ x and
∂ y \partial y ∂ y turns each term into
∂ f / ∂ u \partial f/\partial u ∂ f / ∂ u and doubles the answer; partial derivatives do not cancel like fractions.
❌ That is ∂ f / ∂ v \partial f/\partial v ∂ f / ∂ v The paths ending at
v v v use
x v = 1 x_v = 1 x v = 1 and
y v = − 1 y_v = -1 y v = − 1 ; for
u u u both rates are
1 1 1 .
Show solution
At ( u , v ) = ( 2 , 1 ) (u, v) = (2, 1) ( u , v ) = ( 2 , 1 ) x = 3 x = 3 x = 3 y = 1 y = 1 y = 1 f x = y = 1 f_x = y = 1 f x = y = 1 f y = x = 3 f_y = x = 3 f y = x = 3 x u = 1 x_u = 1 x u = 1 y u = 1 y_u = 1 y u = 1
∂ f ∂ u = ∂ f ∂ x ∂ x ∂ u + ∂ f ∂ y ∂ y ∂ u = ( 1 ) ( 1 ) + ( 3 ) ( 1 ) = 4. \frac{\partial f}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u} = (1)(1) + (3)(1) = 4. ∂ u ∂ f = ∂ x ∂ f ∂ u ∂ x + ∂ y ∂ f ∂ u ∂ y = ( 1 ) ( 1 ) + ( 3 ) ( 1 ) = 4. Check by substitution: f = ( u + v ) ( u − v ) = u 2 − v 2 f = (u + v)(u - v) = u^2 - v^2 f = ( u + v ) ( u − v ) = u 2 − v 2 f u = 2 u = 4 f_u = 2u = 4 f u = 2 u = 4 ∂ f ∂ u + ∂ f ∂ u \frac{\partial f}{\partial u} + \frac{\partial f}{\partial u} ∂ u ∂ f + ∂ u ∂ f 8 8 8
Problem 3 · Both Polar Partials at a Point
Given: f ( x , y ) = x 2 y f(x, y) = x^2 y f ( x , y ) = x 2 y r = 2 r = 2 r = 2 θ = π 6 \theta = \tfrac{\pi}{6} θ = 6 π find f r f_r f r f θ f_\theta f θ
What is f r f_r f r
a f r = 5 3 2 f_r = \tfrac{5\sqrt{3}}{2} f r = 2 5 3 b f r = 2 3 + 3 f_r = 2\sqrt{3} + 3 f r = 2 3 + 3 c f r = 3 f_r = \sqrt{3} f r = 3 d f r = 9 2 f_r = \tfrac{9}{2} f r = 2 9
What is f θ f_\theta f θ
a f θ = 3 f_\theta = \sqrt{3} f θ = 3 b f θ = 3 2 f_\theta = \tfrac{\sqrt{3}}{2} f θ = 2 3 c f θ = 5 3 f_\theta = 5\sqrt{3} f θ = 5 3 d f θ = 3 f_\theta = 3 f θ = 3
✅ Correct! f r = 9 2 f_r = \tfrac{9}{2} f r = 2 9 and
f θ = 3 f_\theta = \sqrt{3} f θ = 3 , both confirmed by differentiating
f = r 3 cos 2 θ sin θ f = r^3\cos^2\theta\sin\theta f = r 3 cos 2 θ sin θ .
❌ Check f r f_r f r At this point
f x = 2 x y = 2 3 f_x = 2xy = 2\sqrt{3} f x = 2 x y = 2 3 and
f y = x 2 = 3 f_y = x^2 = 3 f y = x 2 = 3 ; weight them by
cos θ = 3 2 \cos\theta = \tfrac{\sqrt{3}}{2} cos θ = 2 3 and
sin θ = 1 2 \sin\theta = \tfrac{1}{2} sin θ = 2 1 .
❌ Rates swapped. f x f_x f x goes with
x r = cos θ x_r = \cos\theta x r = cos θ , not
sin θ \sin\theta sin θ .
❌ Check f θ f_\theta f θ Use
x θ = − r sin θ = − 1 x_\theta = -r\sin\theta = -1 x θ = − r sin θ = − 1 and
y θ = r cos θ = 3 y_\theta = r\cos\theta = \sqrt{3} y θ = r cos θ = 3 .
❌ The factor r r r Around a circle of radius
r = 2 r = 2 r = 2 the point moves at speed
r r r :
x θ = − r sin θ x_\theta = -r\sin\theta x θ = − r sin θ ,
y θ = r cos θ y_\theta = r\cos\theta y θ = r cos θ .
❌ Sign of x θ x_\theta x θ Increasing
θ \theta θ in the first quadrant moves the point left, so
x θ = − r sin θ x_\theta = -r\sin\theta x θ = − r sin θ is negative.
Show solution
At r = 2 r = 2 r = 2 θ = π 6 \theta = \tfrac{\pi}{6} θ = 6 π x = 2 cos π 6 = 3 x = 2\cos\tfrac{\pi}{6} = \sqrt{3} x = 2 cos 6 π = 3 y = 2 sin π 6 = 1 y = 2\sin\tfrac{\pi}{6} = 1 y = 2 sin 6 π = 1
Partials of f = x 2 y f = x^2 y f = x 2 y f x = 2 x y = 2 3 f_x = 2xy = 2\sqrt{3} f x = 2 x y = 2 3 f y = x 2 = 3 f_y = x^2 = 3 f y = x 2 = 3
f r = cos θ f x + sin θ f y = 3 2 ( 2 3 ) + 1 2 ( 3 ) = 3 + 3 2 = 9 2 f_r = \cos\theta\, f_x + \sin\theta\, f_y = \tfrac{\sqrt{3}}{2}(2\sqrt{3}) + \tfrac{1}{2}(3) = 3 + \tfrac{3}{2} = \tfrac{9}{2} f r = cos θ f x + sin θ f y = 2 3 ( 2 3 ) + 2 1 ( 3 ) = 3 + 2 3 = 2 9 f θ = − r sin θ f x + r cos θ f y = ( − 1 ) ( 2 3 ) + ( 3 ) ( 3 ) = 3 f_\theta = -r\sin\theta\, f_x + r\cos\theta\, f_y = (-1)(2\sqrt{3}) + (\sqrt{3})(3) = \sqrt{3} f θ = − r sin θ f x + r cos θ f y = ( − 1 ) ( 2 3 ) + ( 3 ) ( 3 ) = 3 Check by substitution: f = r 3 cos 2 θ sin θ f = r^3\cos^2\theta\sin\theta f = r 3 cos 2 θ sin θ f r = 3 r 2 cos 2 θ sin θ = 12 ⋅ 3 4 ⋅ 1 2 = 9 2 f_r = 3r^2\cos^2\theta\sin\theta = 12\cdot\tfrac{3}{4}\cdot\tfrac{1}{2} = \tfrac{9}{2} f r = 3 r 2 cos 2 θ sin θ = 12 ⋅ 4 3 ⋅ 2 1 = 2 9 f θ = r 3 ( cos 3 θ − 2 cos θ sin 2 θ ) = 8 ( 3 3 8 − 3 4 ) = 3 f_\theta = r^3(\cos^3\theta - 2\cos\theta\sin^2\theta) = 8\left(\tfrac{3\sqrt{3}}{8} - \tfrac{\sqrt{3}}{4}\right) = \sqrt{3} f θ = r 3 ( cos 3 θ − 2 cos θ sin 2 θ ) = 8 ( 8 3 3 − 4 3 ) = 3
Problem 4 · From Polar Rates Back to f x f_x f x f y f_y f y
Given: at the point ( x , y ) = ( 0 , 2 ) (x, y) = (0, 2) ( x , y ) = ( 0 , 2 ) f r = 3 f_r = 3 f r = 3 f θ = 4 f_\theta = 4 f θ = 4 find f x f_x f x f y f_y f y
What is f x f_x f x
a f x = 2 f_x = 2 f x = 2 b f x = − 2 f_x = -2 f x = − 2 c f x = − 8 f_x = -8 f x = − 8 d f x = − 4 f_x = -4 f x = − 4
What is f y f_y f y
a f y = 4 f_y = 4 f y = 4 b f y = 3 2 f_y = \tfrac{3}{2} f y = 2 3 c f y = 3 f_y = 3 f y = 3 d f y = 6 f_y = 6 f y = 6
✅ Correct! At
θ = π 2 \theta = \tfrac{\pi}{2} θ = 2 π the relations reduce to
f r = f y f_r = f_y f r = f y and
f θ = − 2 f x f_\theta = -2f_x f θ = − 2 f x , so
f x = − 2 f_x = -2 f x = − 2 and
f y = 3 f_y = 3 f y = 3 .
❌ Check f x f_x f x At
( 0 , 2 ) (0, 2) ( 0 , 2 ) ,
r = 2 r = 2 r = 2 and
θ = π 2 \theta = \tfrac{\pi}{2} θ = 2 π , so
f θ = − r sin θ f x + r cos θ f y = − 2 f x f_\theta = -r\sin\theta\, f_x + r\cos\theta\, f_y = -2f_x f θ = − r sin θ f x + r cos θ f y = − 2 f x .
❌ Sign. x θ = − r sin θ = − 2 x_\theta = -r\sin\theta = -2 x θ = − r sin θ = − 2 here: increasing
θ \theta θ from the top of the circle moves the point left, so
4 = − 2 f x 4 = -2f_x 4 = − 2 f x .
❌ Multiplied instead of divided. f θ = − 2 f x f_\theta = -2f_x f θ = − 2 f x , so
f x = f θ / ( − 2 ) f_x = f_\theta / (-2) f x = f θ / ( − 2 ) , not
− 2 f θ -2f_\theta − 2 f θ .
❌ The factor r r r x θ = − r sin θ x_\theta = -r\sin\theta x θ = − r sin θ carries
r = 2 r = 2 r = 2 , so
f θ = − 2 f x f_\theta = -2f_x f θ = − 2 f x .
❌ Check f y f_y f y With
cos θ = 0 \cos\theta = 0 cos θ = 0 and
sin θ = 1 \sin\theta = 1 sin θ = 1 ,
f r = cos θ f x + sin θ f y f_r = \cos\theta\, f_x + \sin\theta\, f_y f r = cos θ f x + sin θ f y is just
f y f_y f y .
Show solution
The point ( 0 , 2 ) (0, 2) ( 0 , 2 ) r = 2 r = 2 r = 2 θ = π 2 \theta = \tfrac{\pi}{2} θ = 2 π cos θ = 0 \cos\theta = 0 cos θ = 0 sin θ = 1 \sin\theta = 1 sin θ = 1
f r = cos θ f x + sin θ f y = f y ⇒ f y = 3 f_r = \cos\theta\, f_x + \sin\theta\, f_y = f_y \quad\Rightarrow\quad f_y = 3 f r = cos θ f x + sin θ f y = f y ⇒ f y = 3 f θ = − r sin θ f x + r cos θ f y = − 2 f x ⇒ f x = 4 − 2 = − 2 f_\theta = -r\sin\theta\, f_x + r\cos\theta\, f_y = -2f_x \quad\Rightarrow\quad f_x = \frac{4}{-2} = -2 f θ = − r sin θ f x + r cos θ f y = − 2 f x ⇒ f x = − 2 4 = − 2 Check: with f x = − 2 f_x = -2 f x = − 2 f y = 3 f_y = 3 f y = 3 f r = 0 ⋅ ( − 2 ) + 1 ⋅ 3 = 3 f_r = 0\cdot(-2) + 1\cdot 3 = 3 f r = 0 ⋅ ( − 2 ) + 1 ⋅ 3 = 3 f θ = − 2 ⋅ ( − 2 ) + 0 ⋅ 3 = 4 f_\theta = -2\cdot(-2) + 0\cdot 3 = 4 f θ = − 2 ⋅ ( − 2 ) + 0 ⋅ 3 = 4
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