Differentiating equations in x and y implicitly, including products such as x²y³ and terms such as eʸ, ln y and sin y

Tangents and normals to implicitly defined curves, including where a tangent meets the curve again

Stationary points (horizontal tangents) and vertical tangents on implicit curves

Differentiating aˣ by taking logarithms

Forming simple differential equations from a description in words

Connected rates of change, such as dV/dt=dV/dr×dr/dt

Combining all of these in a longer multi-part question

1. Implicit differentiation is the chain rule applied to every term containing y: you differentiate both sides of an equation such as x²+xy+y²=12 with respect to x, treating y as a function of x, and then rearrange for dy/dx.

2. Chain rule on y terms: d/dx[g(y)]=g′(y)dy/dx, so y³→3y²dy/dx and e²ʸ→2e²ʸdy/dx. Product rule on mixed terms: d/dx(xy)=y+xdy/dx. A constant differentiates to 0.

3. The gradient usually contains both x and y, so substitute both coordinates of the point.

4. Horizontal tangent: numerator of dy/dx is zero. Vertical tangent: denominator is zero. Either way, solve together with the curve's equation.

5. d/dx(aˣ)=aˣln a, from ln y = xln a.

6. "Rate of change" means d/dt. Link rates with the chain rule, e.g. dV/dt=dV/dh×dh/dt. "Proportional to" becomes "=k×", with a minus sign for a decrease.

An explicit equation gives y directly in terms of x: y=x³-4x or y=e²ˣsin x. You differentiate these with the P3 rules. An implicit equation mixes x and y without y being the subject: x²+y²=25, x²y+y³=10, xeʸ=2x+y.

Rearranging is often impossible (try making y the subject of xeʸ=2x+y), or it gives two branches, such as y=±√(25-x²). Implicit differentiation avoids rearranging altogether. The idea is that near any point where the curve is not vertical, the curve is the graph of some function y(x), even if you cannot write that function down. So you can differentiate both sides of the equation with respect to x, and the two sides stay equal.

The gradient you get is a formula in both x and y. That is not a flaw; it is necessary.

On the circle x²+y²=25 the points (3,4) and (3,-4) share an x-coordinate but have different gradients, so no formula in x alone could describe both. Differentiating gives 2x+2ydy/dx=0, so dy/dx=-x/y: that is -¾ at (3,4) and +¾ at (3,-4).

Differentiate it as if the letter were x, then multiply by dy/dx. This is the chain rule with y as the inner function. Terms in x only are differentiated normally, and constants give 0.

Mixed terms such as xy or x²y³ are products of a function of x and a function of y, so they need the product rule (which is in the booklet) with the chain rule on the y part.

The method is always the same four steps: (1) differentiate every term on both sides; (2) collect the terms containing dy/dx on one side and everything else on the other; (3) factorise out dy/dx; (4) divide.

WORKED EXAMPLE · QUESTION

The circle C has equation x²+y²-6x+4y-12=0. (a) Find dy/dx in terms of x and y. (b) Find an equation of the tangent to C at P(7,1), in the form ax+by+c=0 where a, b and c are integers.

SOLUTION

(a) Differentiate term by term: 2x+2ydy/dx-6+4dy/dx=0. Collect and factorise: (2y+4)dy/dx=6-2x, so dy/dx=(3-x)/(y+2).

(b) Check P is on C: 49+1-42+4-12=0 ✓. Gradient at P: (3-7)/(1+2)=-(4/3). Tangent: y-1=-(4/3)(x-7), so 3y-3=-4x+28, giving 4x+3y-31=0.

Check by geometry: completing the square gives centre (3,-2) and radius 5. The radius to P has gradient ¾, and ¾×(-(4/3))=-1, so the tangent is perpendicular to the radius, as it must be.

WORKED EXAMPLE · QUESTION

(a) The curve C₁ has equation x³-2xy+y²=5. Find the gradient of C₁ at the point (2,3). (b) The curve C₂ has equation xeʸ=2x+y. Find dy/dx in terms of x and y, and its value at the origin.

SOLUTION

(a) Product rule on 2xy (in a bracket because of the minus sign), chain rule on y²: 3x²-(2y+2xdy/dx)+2ydy/dx=0. Collect: (2y-2x)dy/dx=2y-3x², so dy/dx=(2y-3x²)/(2y-2x). The point is on C₁ since 8-12+9=5. At (2,3): (6-12)/(6-4)= -3.

(b) Product rule on xeʸ; the right-hand side gives 2+dy/dx: eʸ+xeʸdy/dx=2+dy/dx. So (xeʸ-1)dy/dx=2-eʸ and dy/dx=(2-eʸ)/(xeʸ-1). At (0,0): (2-1)/(0-1)= -1.

The method is the one from P1: find the gradient m at the point, use -1/m for the normal, then y-y₁=m(x-x₁). The only change is that dy/dx depends on x and y, so you need both coordinates of the point. If you are given only x, substitute it into the curve's equation and solve for y first; there may be more than one point.

To find where a line meets the curve again, substitute the line into the curve's equation. A tangent gives a repeated root at the point of contact, because the line touches the curve there. Knowing this root twice over makes the resulting cubic easy to factorise, and it is also a strong check that your tangent is right.

WORKED EXAMPLE · QUESTION

The curve C has equation x²y+y³=10. The point P(3,1) lies on C. (a) Find dy/dx in terms of x and y. (b) Find an equation of the tangent to C at P. (c) The tangent at P meets C again at Q. Find the coordinates of Q. (d) Find an equation of the normal to C at P.

SOLUTION

(a) Product rule on x²y, chain rule on y³: 2xy+x²dy/dx+3y²dy/dx=0, so dy/dx=(-2xy)/(x²+3y²).

(b) At P(3,1): dy/dx=(-6)/(9+3)=-½. Then y-1=-½(x-3), so x+2y-5=0.

(c) Substitute x=5-2y into C: (5-2y)²y+y³=10, which expands to 5y³-20y²+25y-10=0, i.e. y³-4y²+5y-2=0. The tangent touches at P, so y=1 is a repeated root: (y-1)²(y-2)=0. Hence y=2 and x=5-4=1: Q(1,2). Check: 1×2+8=10 ✓.

(d) Normal gradient =-1÷(-½)=2. So y-1=2(x-3), i.e. y=2x-5.

A stationary point has dy/dx=0: the tangent is horizontal. For a fraction this means the numerator is zero (with the denominator not zero). Setting the numerator to zero gives a relationship between x and y, such as y=-2x, not the points themselves. Geometrically it is a whole line of possible points; the actual stationary points are where that line meets the curve, so you substitute it into the curve's equation.

A vertical tangent has dx/dy=0. Since dx/dy=1/dy/dx (P3), this happens where the denominator of dy/dx is zero. Same method: denominator =0, then substitute into the curve. In words, "parallel to the x-axis" means horizontal and "parallel to the y-axis" means vertical. To decide which stationary point is a maximum and which a minimum, the shape of the curve is usually enough: a closed curve has its highest point at the stationary point with the larger y.

WORKED EXAMPLE · QUESTION

The curve C has equation x²+xy+y²=12. (a) Show that dy/dx=-(2x+y)/(x+2y). (b) Find the coordinates of the points on C at which the tangent is parallel to the x-axis. (c) Find the coordinates of the points on C at which the tangent is parallel to the y-axis.

SOLUTION

(a) Product rule on xy: 2x+y+xdy/dx+2ydy/dx=0, so (x+2y)dy/dx=-(2x+y) and dy/dx=-(2x+y)/(x+2y) as required.

(b) Numerator zero: 2x+y=0, so y=-2x. Substitute: x²-2x²+4x²=12, so 3x²=12 and x=±2. The points are (2,-4) and (-2,4). Check the denominator: at (2,-4), x+2y=-6≠0.

(c) Denominator zero: x=-2y. Substitute: 4y²-2y²+y²=12, so y=±2. The points are (-4,2) and (4,-2).

The curve is a closed oval, so (-2,4) is its highest point and (2,-4) its lowest.

The power rule does not apply to 3ˣ: it needs a constant power and a variable base, and here the base is constant while the power varies. Instead take natural logarithms, which turns the power into a product, and then differentiate implicitly.

When a=e, ln e=1 and the rule gives back d/dx(eˣ)=eˣ. For 0<a<1, ln a<0 and the gradient is negative everywhere. Notice that the gradient aˣln a is a constant multiple of the height aˣ: every exponential curve grows at a rate proportional to its own value, which is exactly why exponentials model growth.

WORKED EXAMPLE · QUESTION

(a) By taking natural logarithms, show that if y=aˣ, where a is a positive constant, then dy/dx=aˣln a. (b) The tangent to y=3ˣ at the point (2,9) meets the x-axis at A. Find the exact x-coordinate of A. (c) Differentiate y=4³ˣ.

SOLUTION

(a) ln y=xln a, so 1/ydy/dx=ln a, and dy/dx=yln a=aˣln a as required.

(b) Gradient at x=2: 3²ln 3=9ln 3 (≈9.89). Tangent: y-9=9ln 3 (x-2). Put y=0: -9=9ln 3 (x-2), so x=2-1/ln 3 (≈1.09).

(c) ln y=3xln 4, so 1/ydy/dx=3ln 4 and dy/dx=3ln 4·4³ˣ (=6ln 2·4³ˣ).

A differential equation is an equation containing a derivative. In context, "the rate of change of N" (with time) is dN/dt. The translation from words follows a few fixed rules:

"A is proportional to B" becomes A=kB, where k is a constant of proportionality; "inversely proportional to B" gives A=k/B.

Take k>0 and put the sign in yourself: a quantity that decreases gets a minus sign, dm/dt=-km.

If a rate is given at one moment ("when θ=78 it is falling at 2.4 degrees per minute"), substitute it to find k. "Falling at 2.4" means the derivative is -2.4.

Several effects at once add up: "water flows in at 20 cm³ s⁻¹ and leaks out at a rate proportional to the depth" gives dV/dt=20-kh.

WORKED EXAMPLE · QUESTION

Write a differential equation for each situation, stating the meaning of any constant. (a) The number of fish, N, in a lake t years after stocking increases at a rate proportional to N. (b) The depth, h m, of water in a draining tank decreases at a rate proportional to the square root of h. (c) A drink at temperature θ °C stands in a room at 18 °C. Its temperature falls at a rate proportional to the difference between its temperature and the room temperature. When θ=78 it is cooling at 2.4 °C per minute. Find the constant, and the rate of cooling when θ=48.

SOLUTION

(a) dN/dt=kN, where k>0 is a constant.

(b) Decreasing, so a minus sign: dh/dt=-k√h, k>0.

(c) The difference is θ-18 and the temperature is falling: dθ/dt=-k(θ-18). Substitute: -2.4=-k(78-18), so k=0.04. At θ=48: dθ/dt=-0.04×30=-1.2, so the drink is cooling at 1.2 °C per minute.

Here t is in minutes, so k has units per minute. Solving these equations, to get N, h or θ as functions of t, is done on the P4 differential equations page.

Often you know one rate (how fast air is pumped in, dV/dt) and want another (how fast the radius grows, dr/dt). A geometric formula links the quantities, V=(4/3)πr³, so you can find dV/dr. The chain rule joins them:

Method: (1) write the formula linking the two quantities; (2) differentiate it; (3) write the chain rule and substitute the known rate and the value at that instant; (4) solve, with units. Substitute the particular value last: if you put r=10 into V before differentiating, V becomes a constant and its derivative is 0.

A first example: a stone dropped in a pond makes a circular ripple whose radius grows at a steady 3 cm s⁻¹. The area is A=πr², so dA/dt=2πr×3=6πr. When r=15 cm the area is growing at 90π≈283 cm² s⁻¹. The radius grows at a constant rate, but the area grows faster and faster.

WORKED EXAMPLE · QUESTION

A spherical balloon is inflated so that its volume increases at a constant rate of 50 cm³ s⁻¹. [The volume of a sphere of radius r is (4/3)πr³ and its surface area is 4πr².] When the radius is 10 cm, find (a) the rate at which the radius is increasing, (b) the rate at which the surface area is increasing.

SOLUTION

(a) V=(4/3)πr³, so dV/dr=4πr²=400π when r=10. Chain rule: dV/dt=dV/dr×dr/dt, so 50=400π×dr/dt and dr/dt=1/8π≈0.0398 cm s⁻¹.

(b) S=4πr², so dS/dr=8πr=80π. Then dS/dt=80π×1/8π= 10 cm² s⁻¹.

Keep dr/dt exact for part (b): the π cancels and the answer is exact.

Longer questions combine a geometric formula (often with similar triangles), a connected-rates calculation and a differential equation formed from a description. The next example does all three. The key geometric step is that in a cone held vertex down, the water always forms a smaller cone of the same shape, so its radius is a fixed fraction of its depth and the volume can be written in terms of the depth alone.

WORKED EXAMPLE · QUESTION

A container is a hollow right circular cone of height 12 cm and top radius 6 cm, held with its axis vertical and its vertex at the bottom. Water is poured in at a constant rate of 20 cm³ s⁻¹. When the depth of water is h cm the volume of water is V cm³. [The volume of a cone is ⅓πr²h.] (a) Show that V=πh³/12. (b) Assuming no water is lost, find the rate at which the depth is increasing when h=8. In a refined model, water also leaks from a small hole at the vertex at a rate of kh cm³ s⁻¹, where k is a positive constant. (c) Show that dh/dt=(4(20-kh))/πh². (d) In this model the depth stays constant when h=10. Find k. (e) Find the rate at which the depth is increasing when h=8 in the refined model, and compare it with (b).

SOLUTION

(a) Similar triangles: r/h=(6/12), so r=h/2. Then V=⅓π(h/2)²h=πh³/12 as required.

(b) dV/dh=πh²/4=16π at h=8. So dh/dt=dV/dt÷dV/dh=20/16π= 5/4π≈0.398 cm s⁻¹.

(c) Net rate in = rate in − rate out: dV/dt=20-kh. Also dV/dt=πh²/4dh/dt.

Equate: πh²/4dh/dt=20-kh, so dh/dt=(4(20-kh))/πh² as required.

(d) Depth constant means dh/dt=0: 20-10k=0, so k=2.

(e) With k=2, h=8: dh/dt=(4(20-16))/64π= 1/4π≈0.0796 cm s⁻¹. This is one fifth of the answer to (b): the leak removes 16 of the 20 cm³ entering each second at this depth, so the water rises much more slowly. In this model the depth approaches 10 cm and the cone (12 cm tall) never overflows.

P3 differentiation

Every step here is the chain and product rules from P3; dy/dx=1/dx/dy explains why a zero denominator means a vertical tangent, and aˣ=e^(xln a) gives a second route to the derivative of aˣ.

P1 and P2 coordinate geometry

Tangents and normals use y-y₁=m(x-x₁) and perpendicular gradients from P1; circles from P2 give a geometric check (tangent perpendicular to radius).

Other P4 pages

P4.2 Parametric equations give dy/dx by a different route; P4.7 Differential equations solves the equations formed in this page, and P4.6 Volumes of revolution builds on the volume formulae used for rates. In Mechanics, rates of change in time are velocity and acceleration, and dN/dt=-λN (radioactive decay) and Newton's law of cooling are the physics versions of forming differential equations.

1. Find dy/dx for the curve y²=4x.

2. Find the gradient of the curve xy=6 at the point (2,3).

3. What is the gradient of y=2ˣ where it crosses the y-axis?

4. The radius of a circle increases at ½ cm s⁻¹. How fast is the area increasing when the radius is 5 cm?

5. Write a differential equation for: "the mass m of a melting block of ice decreases at a rate inversely proportional to m".

6. In worked example 4, show that the point (-2,4) is on the curve and confirm the tangent there is horizontal.

1. 2ydy/dx=4, so dy/dx=2/y. For example, at (1,2) the gradient is 1 and at (1,-2) it is -1.

2. y+xdy/dx=0, so dy/dx=-y/x=-(3/2).

3. dy/dx=2ˣln 2, which is ln 2≈0.693 at x=0.

4. dA/dt=2πr×dr/dt=2π×5×½=5π≈15.7 cm² s⁻¹.

5. dm/dt=-k/m, where k is a positive constant. "Inversely proportional" puts m in the denominator; "decreases" gives the minus sign.

6. 4+(-2)(4)+16=12 ✓. The numerator 2x+y=-4+4=0 and the denominator x+2y=-2+8=6≠0, so dy/dx=0: the tangent is horizontal.