Teacher page: implicit differentiation and related rates (AA 5.14)

The one thing to do with the figure

Ask for the gradient at the top of the circle.

At (0, 5) the tangent is horizontal and −x/y gives 0. At (5, 0) it is vertical and the formula divides by zero, which is the honest answer rather than an error.

Moving the point also shows the tangent staying perpendicular to the radius throughout, which is what −x/y encodes.

Every y produces a dy/dx

Differentiating y³ with respect to x gives 3y² dy/dx, not 3y². The chain rule is doing it, and the extra factor is the entire method.

A student who misses one of them gets an equation that cannot be rearranged, and usually blames the algebra.

The answers

x² + y² = 25	2x + 2y dy/dx = 0, so dy/dx = −x/y.
At (3, 4)	Gradient −¾, tangent 3x + 4y = 25.
The ladder	dy/dt = −(x/y)(dx/dt) = −¾ × 0.5 = −0.375 m per s.
1, 2, 3	−4/3; B, 3y² dy/dx; −0.375.

Where the marks go

1 markProducing a dy/dx for every y term.

1 markCollecting the dy/dx terms and making it the subject.

1 markOn related rates, differentiating with respect to t before substituting.

1 markInterpreting the sign in words.

What each wrong answer tells you

They give	What it means
3y² with no dy/dx	The defining error. Every y is a function of x.
Substituting before differentiating	Turns a variable into a constant and its rate vanishes. Produces zero or nonsense.
+0.375 for the ladder	Sign. The top descends as the foot slides out.
−0.75 for the ladder	Gave dy/dx rather than dy/dt. The 0.5 was never used.

Other things they will say

"Why is y a function of x?" Locally it is: on the top half of the circle, y really is determined by x. Implicit differentiation works on that half and the bottom half simultaneously.

"Can I just rearrange?" For a circle, yes, into two branches with a root each. For x³ + y³ = 6xy you cannot, which is why the method exists.

"The answer has y in it." It usually does, and that is why you need a point rather than just an x value.

A possible order

	What is happening
1	Show a circle and ask for y = f(x). Let the vertical line test do the work.
2	The method, with every dy/dx written in a different colour.
3	The animation: tangent and radius perpendicular, and the vertical case.
4	Related rates, with respect to t, numbers last.
5	Questions 1 to 3.
6	Products like xy, as a preview.

Two things not to say

Do not substitute the point before differentiating, even to save time. Students copy it and the method breaks silently.

Do not let a related rates answer stop at a number. The sign means something, and saying what is usually the last mark.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

A circleFor x² + y² = 25, find dy/dx and evaluate at (3, 4).
2x + 2y(dy/dx) = 0, so dy/dx = −x/y = −0.75.
The normalFind the gradient of the normal there.
4/3 = 1.333

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

A harder curveFor x² + xy + y² = 7, find dy/dx at (1, 2).
2x + y + x(dy/dx) + 2y(dy/dx) = 0, so dy/dx = −(2x + y)/(x + 2y) = −4/5 = −0.8. The point checks: 1 + 2 + 4 = 7.
Why the chain rule appearsExplain why differentiating y² with respect to x gives 2y(dy/dx).
y is a function of x, so y² is a function of a function. The chain rule supplies the dy/dx factor. Writing 2y alone is the standard error.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

A geometric checkThe normal to a circle at any point passes through its centre. Verify that at (3, 4) for x² + y² = 25.
The normal gradient is 4/3 and the line through (3, 4) with that gradient is y = 4x/3, which passes through the origin, the centre. A result that can be checked geometrically is worth checking.
When y cannot be isolatedExplain why implicit differentiation is necessary rather than merely convenient for x³ + y³ = 6xy.
That relation cannot be rearranged into y as an elementary function of x, so there is no explicit form to differentiate. Implicit differentiation is the only route, and it also handles curves that fail the vertical line test.

Practicalities

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