Implicit differentiation and related rates: the one idea behind both, worked on a curve and on the ladder problem

Implicit differentiation and related rates are the same move. In both, you differentiate an equation whose variables depend on something else, and every time you differentiate a variable that is not the one you are differentiating with respect to, the chain rule hands you a factor: y' when you differentiate y with respect to x, dx/dt when you differentiate x with respect to time. Forget the factor and both methods collapse. Here is one curve and the ladder problem, each worked line by line.

The one idea: a hidden chain rule in implicit differentiation

When y is a function of x, the expression y² is a function inside a function. The outside is the square. The inside is y. So the chain rule applies: d/dx (y²) = 2y · y', where y' is dy/dx. It is not 2y, because y changes when x changes.

Now swap the letters. When x is a function of time t, the expression x² is again a function inside a function, and d/dt (x²) = 2x · dx/dt. Same rule, different letters. Implicit differentiation is this rule with x as the variable you differentiate by. Related rates is this rule with t.

That is the whole page in one habit: every time you differentiate a y, multiply by y'; every time you differentiate with respect to t, multiply by that variable's rate. If the chain rule is solid, this is not a new technique. It is the chain rule applied to a letter instead of a bracket.

Worked example 1: implicit differentiation on a curve, line by line

Find dy/dx for the curve x² + xy + y² = 7 at the point (1, 2).

First, check the point is on the curve. Substitute x = 1 and y = 2: 1 + 2 + 4 = 7. It is. A point that is not on the curve gives a slope that means nothing, so this check comes first.

• Line 1, differentiate x². The derivative with respect to x is 2x. No hidden factor, because x is the variable.
• Line 2, differentiate xy with the product rule. Both factors change. The derivative is 1 · y + x · y' = y + x y'.
• Line 3, differentiate y² with the hidden chain rule. 2y · y'.
• Line 4, differentiate the right side. 7 is a constant, so its derivative is 0.
• Line 5, write the differentiated equation. 2x + y + x y' + 2y y' = 0.
• Line 6, collect the y' terms and factor. y'(x + 2y) = −(2x + y), so y' = −(2x + y)/(x + 2y).
• Line 7, substitute the point last. At (1, 2): y' = −(2 + 2)/(1 + 4) = −4/5.

The slope of the curve at (1, 2) is −4/5.

The common wrong lines

d/dx (y²) = 2y, the y' dropped. This is the error the page exists for. It treats y as a constant. But y moves as x moves, and the chain rule's inner derivative is exactly y'. Without it, line 6 has nothing to collect, and the answer is wrong.

d/dx (xy) = y' alone, the product rule skipped. Both x and y change, so both get differentiated in turn. The y in "y + x y'" is the part that goes missing.

Substituting (1, 2) before differentiating. If you put the numbers in first, every term becomes a constant and every derivative becomes 0. The equation is differentiated with its variables still in it. The point goes in on the last line.

Related rates: the same idea with t

Related rates problems describe two or more quantities that change over time and are linked by an equation. You know how fast one changes; you want how fast another changes at one particular instant. Here are the five steps, in order.

1. Draw and label. Name every changing quantity with a letter. A fixed length gets its number.
2. Write the equation that relates them. It must be true at every instant, not just the one in the question.
3. Differentiate with respect to t. Every changing variable picks up its own rate, such as dx/dt or dy/dt.
4. Substitute the instant's values and the known rates. Only now do the numbers go in.
5. Solve for the wanted rate, and say its sign in words.

Step 3 is the hidden chain rule from the first section. Step 4 is the "substitute last" rule from the curve.

Worked example 2: the ladder problem, line by line

A 10 m ladder leans against a vertical wall. Its foot slides away from the wall at 0.5 m/s. How fast is the top sliding down the wall when the foot is 6 m from the wall?

• Line 1, label and relate. Let x be the distance from the wall to the foot and y the height of the top, both in metres. The wall, the ground and the ladder make a right triangle, so x² + y² = 100 at every instant.
• Line 2, differentiate with respect to t. 2x · dx/dt + 2y · dy/dt = 0. Each variable carries its rate.
• Line 3, find the instant's values. x = 6, so y² = 100 − 36 = 64 and y = 8.
• Line 4, the known rate. dx/dt = 0.5 m/s, positive because the foot moves away from the wall.
• Line 5, substitute and solve. 2(6)(0.5) + 2(8) · dy/dt = 0, so 6 + 16 · dy/dt = 0, so dy/dt = −6/16 = −3/8 = −0.375 m/s.
• Line 6, say it in words. The top of the ladder is sliding down the wall at 0.375 m/s at that instant.

The common wrong lines

Substituting x = 6 and y = 8 before differentiating. Then the equation reads 36 + 64 = 100, its derivative is 0 = 0, and nothing is left to solve. The relation must still contain the variables when it is differentiated.

dy/dt = −0.5 "because the ladder is rigid". The ladder's length is fixed, but its two ends do not move at the same speed. Here the top moves at 0.375 m/s while the foot moves at 0.5 m/s. The geometry decides, through line 5.

Dropping the minus and reporting the top rising. A positive dy/dt would mean the height is growing. Read the sign before you write the sentence.

Siblings to try cold

Cover both worked examples. Do each one on paper, every line, before you look below.

• Implicit: find dy/dx for x³ + y³ = 9 at the point (1, 2). Check the point is on the curve first.
• Related rates: a 13 m ladder leans against a wall. Its foot slides away at 0.2 m/s. How fast is the top sliding down when the foot is 5 m from the wall?

Check

• The point checks: 1 + 8 = 9. Then y' = −x²/y², and at (1, 2) that is −1/4.
• When x = 5, y = 12. Then dy/dt = −(5/12)(0.2) = −1/12 m/s, about −0.083 m/s. In words: the top slides down at about 0.083 m/s.

If either went wrong, find the line where it went wrong. A missing y', or numbers substituted too early, accounts for most misses, and that line is tonight's card.

Reading the sign

A negative rate means the quantity is decreasing. In the ladder problem dy/dt = −0.375 m/s because the height is shrinking. But the question asked how fast the top is sliding down, and "down" already carries the direction. So the answer in words is "sliding down at 0.375 m/s", a positive size with the direction said. Writing "−0.375 m/s down" says the top is going up. Marks are lost on that sentence, not on the arithmetic, so write the sentence as carefully as the derivative.

Where this sits in the course, and where it goes next

In MCV4U, implicit differentiation is idea 9 of 24, Implicit differentiation. Before it is idea 8, Chain rule; after it is idea 10, Derivatives of exponential and logarithmic functions. Related rates is idea 17, Related rates, with idea 16, Optimization problems, before it; after it the vectors strand begins at idea 18. The MCV4U course map shows all 24 in teaching order.

In a first-year Calculus I course, implicit differentiation is idea 7 of 26, Implicit differentiation, between idea 6, The chain rule, and idea 8, Derivatives of exponential and logarithmic functions. Related rates is idea 17, Related rates, after idea 16, Optimization, and before idea 18, Linear approximation and Newton's method. First-year calculus covers what changes from Grade 12, and related rates is a regular in AP Calculus too.

The slope you found on the curve has an immediate use. The tangent line at (1, 2) has slope −4/5, so its equation is y − 2 = −(4/5)(x − 1). That is the most common follow-up question on an implicit curve. If any rule felt shaky along the way, the derivative rules puts them all on one page.

Practise these mixed, not in blocks. A set with only implicit questions lets you skip the decision about which rule a term needs. Rohrer and Taylor (2007) found that students who practised shuffled problem types scored higher on a later test than students who practised the same problems in blocks. Mix the problem types has the method: put implicit and explicit derivatives, and a related rates problem, in one sitting. The order on this page, a worked example first and then a sibling done cold, follows Sweller and Cooper (1985), and worked examples, then practice problems explains why.

Implicit differentiation and related rates on one screen

• The hidden chain rule: d/dx (y²) = 2y · y'; d/dt (x²) = 2x · dx/dt. Same rule, different letters.
• Implicit, in order: differentiate every term, product rule where x and y meet, collect y', factor, divide.
• Related rates, in five steps: draw and label, relate, differentiate with respect to t, substitute, solve and say the sign.
• Substitute last: numbers go in after the derivative, never before.
• The curve: x² + xy + y² = 7 has slope −4/5 at (1, 2).
• The ladder: 10 m, foot at 6 m moving out at 0.5 m/s, top sliding down at 0.375 m/s.