01 · The lower circle branch
Find dy/dx at (3, −4) on x² + y² = 25.
Hint
The derivative formula includes the signed y-coordinate.
Worked solution
dy/dx = −3/(−4) = 3/4. The same x-coordinate as (3, 4) gives the opposite gradient.
Understand · explore · practise
Implicit differentiation
Differentiate equations involving both x and y using the chain and product rules. Find gradients, tangents, normals and horizontal or vertical tangents without solving explicitly for y.
Before you startChain and product rules; tangent and normal equations; real domains
01 / Treat y as a function of x
An equation can describe a curve without isolating y.
d/dx(y²) = 2y(dy/dx)
d/dx(y³) = 3y²(dy/dx)
When differentiating with respect to x, remember that y also changes along the curve. The chain rule supplies a factor dy/dx whenever you differentiate a function of y. We will abbreviate dy/dx to y′.
An implicit curve may have several branches. The derivative is local to a point; two points with the same x-coordinate need not have the same gradient.
One equation, two branchesExplore
2x + 8y y′ = 0.
Point = (0, 2); coefficient of y′ = 16; remaining term = 0; gradient = 0.
The tangent is horizontal because the remaining term is zero and the coefficient of y′ is nonzero.
The same x can give two different y-values and two different gradients. At the left or right endpoint the branches join and the tangent is vertical; a zero denominator is not a finite slope.
02 / Differentiate both sides of a curve equation
Collect the terms containing y′ before dividing.
x² + y² = 25Worked example
2x + 2y y′ = 0
Differentiate every term. The constant derivative is zero.
2y y′ = −2x
Move the term without y′ to the other side.
y′ = −x/y, provided y ≠ 0
Divide by the coefficient of y′.
At (3, 4), y′ = −3/4
First check 3² + 4² = 25.
Watch the local tangent change around an implicit ellipse
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Apply the chain rule to every y-expression
The inside derivative is y′.
d/dx[sin y] = cos y · y′
d/dx[eʸ] = eʸy′
d/dx[ln y] = y′/y, for y > 0
x² + y³ = 9Worked example
2x + 3y²y′ = 0
Differentiate both sides.
y′ = −2x/(3y²), where y ≠ 0
Collect and divide.
At (1, 2): y′ = −1/6
The point satisfies 1 + 8 = 9.
02 · A shifted power of y
Differentiate x + (y + 2)³ = 9.
Hint
The derivative of y + 2 is y′.
Worked solution
1 + 3(y + 2)²y′ = 0, so y′ = −1/[3(y + 2)²] where y ≠ −2.
03 · A sine of y
For x + sin y = 0, find the gradient at (0, 0).
Hint
Differentiate to get 1 + cos y · y′ = 0.
Worked solution
y′ = −1/cos y where cos y ≠ 0. At (0, 0), the gradient is −1.
04 / Use the product rule for mixed x and y terms
Both factors can change with x.
d/dx(xy) = y + xy′
d/dx(x²y) = 2xy + x²y′
x² + xy + y² = 7Worked example
2x + y + xy′ + 2yy′ = 0
The mixed term xy contributes two terms.
(x + 2y)y′ = −(2x + y)
Group all terms containing y′.
y′ = −(2x + y)/(x + 2y)
The formula applies when x + 2y ≠ 0.
At (1, 2): y′ = −4/5
The point satisfies 1 + 2 + 4 = 7.
04 · A mixed polynomial
Find y′ for x²y + y² = 6, and evaluate it at (1, 2).
Hint
Differentiate x²y with the product rule.
Worked solution
2xy + x²y′ + 2yy′ = 0, so y′ = −2xy/(x² + 2y). At (1, 2), it is −4/5.
05 · Another mixed product
Differentiate xy² + x² = 5.
Hint
The derivative of xy² is y² + 2xyy′.
Worked solution
y² + 2xyy′ + 2x = 0, hence y′ = −(y² + 2x)/(2xy) wherever xy ≠ 0.
05 / Keep exponential and logarithmic domains
Differentiate the equation on its original real domain.
eʸ + xy = 1Worked example
eʸy′ + y + xy′ = 0
Use the chain rule and product rule.
(eʸ + x)y′ = −y
Collect the y′ terms.
y′ = −y/(eʸ + x)
Divide only when eʸ + x is nonzero.
At (0, 0): y′ = 0
The point lies on the curve and the denominator is 1.
06 · A logarithmic curve
For ln y + x² = 0, find y′ and its value at (0, 1).
Hint
The equation requires y > 0.
Worked solution
y′/y + 2x = 0, giving y′ = −2xy. At (0, 1), it is zero. The explicit form y = e^(−x²) confirms the derivative.
07 · Exponential product
Find y′ for x eʸ + y = 1.
Hint
The derivative of x eʸ is eʸ + x eʸy′.
Worked solution
(x eʸ + 1)y′ = −eʸ, so y′ = −eʸ/(x eʸ + 1), where the denominator is nonzero.
06 / Choose an efficient way to handle a quotient
Multiplying through preserves the original exclusions.
y/x + x = 3, with x ≠ 0Worked example
y + x² = 3x, still with x ≠ 0
Multiply by x, retaining the original domain.
y′ + 2x = 3
Differentiate the simpler equation.
y′ = 3 − 2x, for x ≠ 0
The original curve still has a missing point at x = 0.
Quotient-rule check: (xy′ − y)/x² + 1 = 0
Substituting y = 3x − x² gives the same result.
08 · A reciprocal y-term
For x + 1/y = 2, find y′ and state the original exclusion.
Hint
The derivative of y⁻¹ is −y⁻²y′.
Worked solution
1 − y′/y² = 0, so y′ = y². The original equation excludes y = 0; equivalently y = 1/(2 − x), with x ≠ 2.
07 / Check the point before evaluating a derivative
A formal expression is not evidence that a point lies on the curve.
Test (1, 1) for x² + xy + y² = 7Worked example
1² + 1·1 + 1² = 3, not 7
This point is not on the stated curve.
Do not report −(2 + 1)/(1 + 2) = −1 as its curve gradient
The derivative formula applies at points satisfying the curve equation.
09 · Solve for both branches first
Find the points with x = 1 on x² + xy + y² = 7, and the gradient at each.
Hint
y² + y − 6 = 0.
Worked solution
y = 2 or −3. At (1, 2), the gradient is −4/5. At (1, −3), it is −1/5. Each calculation uses its own signed y-coordinate.
08 / Write a tangent or normal from the local slope
The usual point–slope method still applies.
x² + xy + y² = 7 at (1, 2)Worked example
Tangent gradient −4/5
Evaluate the implicit derivative.
Tangent: y − 2 = −(4/5)(x − 1)
Use the given contact point.
Normal gradient 5/4
Take the negative reciprocal.
Normal: y − 2 = (5/4)(x − 1)
Check that the line passes through the contact point.
10 · An implicit normal
Find the normal to x² + y³ = 9 at (1, 2).
Hint
The tangent gradient is −1/6.
Worked solution
The normal gradient is 6, so the normal is y − 2 = 6(x − 1).
09 / Find horizontal tangents using the original curve too
A zero numerator must correspond to a valid point.
Horizontal tangents to x² + xy + y² = 7Worked example
2x + y = 0, so y = −2x
Set the numerator of y′ to zero.
x² − 2x² + 4x² = 7 gives 3x² = 7
Substitute into the original curve.
x = ±√(7/3), with y = −2x
These are the two candidate points.
x + 2y = −3x ≠ 0
The derivative denominator is nonzero, so both tangents are horizontal.
11 · An ellipse’s horizontal tangents
Find all horizontal-tangent points on x² + 4y² = 16.
Hint
y′ = −x/(4y). Set x = 0 and solve the original equation.
Worked solution
The points are (0, 2) and (0, −2). The denominator is nonzero at both. Their tangents are y = 2 and y = −2.
12 · Check an impossible condition
Does xy = 4 have any point with a horizontal tangent?
Hint
y′ = −y/x on this curve.
Worked solution
No. A zero derivative would require y = 0, which contradicts xy = 4. Also x = 0 is not on the curve.
10 / Investigate vertical tangents before dividing
A zero y′ coefficient with a nonzero remaining term gives a vertical direction.
Vertical tangents to x² + 4y² = 16Worked example
2x + 8yy′ = 0
Keep the undivided derivative relation.
At y = 0, the curve gives x = ±4
The coefficient of y′ vanishes, while 2x is nonzero.
The tangents are x = 4 and x = −4
Their normals are horizontal. There is no finite dy/dx at either point.
More generally, if the differentiated equation is A + By′ = 0, then y′ = −A/B when B ≠ 0. A regular vertical tangent occurs when B = 0 and A ≠ 0. If A and B both vanish, investigate separately.
13 · A tilted ellipse’s vertical tangents
Find the vertical-tangent points of x² + xy + y² = 7.
Hint
Set x + 2y = 0, then substitute into the original equation.
Worked solution
x = ±√(28/3), and y = −x/2. At either point, 2x + y = 3x/2 is nonzero, so the tangent is vertical.
11 / Do not classify a 0/0 point from the quotient
The curve can have more than one local direction.
x² − y² = 0 at (0, 0)Worked example
2x − 2yy′ = 0 gives y′ = x/y away from y = 0
At the origin the formula is 0/0.
(x − y)(x + y) = 0
The curve consists of two intersecting straight lines.
The branches y = x and y = −x have slopes +1 and −1
There is no single tangent direction to the whole crossing at the origin.
14 · Investigate a cusp
Investigate y² = x³ at the origin.
Hint
The branches are y = ±x^(3/2) for x ≥ 0. Compare their slopes as x approaches zero.
Worked solution
Both branch slopes tend to zero, so the cusp has a horizontal tangent y = 0. The derivative quotient 3x²/(2y) is 0/0 at the origin and cannot be substituted into directly.
15 · Check an implicit derivative
A student differentiates x² + y² = 25 as 2x + 2y = 0. What is missing?
Hint
y is a function of x.
Worked solution
The derivative of y² is 2y y′. The correct equation is 2x + 2y y′ = 0. Omitting y′ loses the chain rule.
12 / Differentiate, collect, check and interpret
The original equation stays part of the solution.
- Treat y as a changing function of x.
- Use chain and product rules wherever needed.
- Collect the terms containing y′, then divide only by a nonzero coefficient.
- Check every proposed point in the original equation.
- For horizontal or vertical tangents, verify the other coefficient is nonzero.
- Investigate singular points and retain domain restrictions.
16 · A mixed final check
For xy + y² = 6, find the tangent at (1, 2).
Hint
The derivative equation is y + (x + 2y)y′ = 0.
Worked solution
The gradient is −2/5, so the tangent is y − 2 = −(2/5)(x − 1). The contact point satisfies 2 + 4 = 6.
Section 1 of 12 · Treat y as a function of x