P4.4 Implicit differentiation and rates of change
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Pure Mathematics 4 (A2), specification sections 5.1–5.2. The chain, product and quotient rules themselves are P3; parametric differentiation and solving differential equations have their own P4 pages.
What this page covers
7 points
- Differentiating equations in \(x\) and \(y\) implicitly, including products such as \(x^2y^3\) and terms such as \(e^y\), \(\ln y\) 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 \(a^x\) by taking logarithms
- Forming simple differential equations from a description in words
- Connected rates of change, such as \(\frac{dV}{dt}=\frac{dV}{dr}\times\frac{dr}{dt}\)
- Combining all of these in a longer multi-part question
In one minute
6 points
- Implicit differentiation is the chain rule applied to every term containing \(y\): you differentiate both sides of an equation such as \(x^2+xy+y^2=12\) with respect to \(x\), treating \(y\) as a function of \(x\), and then rearrange for \(\frac{dy}{dx}\).
- Chain rule on \(y\) terms: \(\frac{d}{dx}[g(y)]=g'(y)\frac{dy}{dx}\), so \(y^3\to 3y^2\frac{dy}{dx}\) and \(e^{2y}\to 2e^{2y}\frac{dy}{dx}\). Product rule on mixed terms: \(\frac{d}{dx}(xy)=y+x\frac{dy}{dx}\). A constant differentiates to 0.
- The gradient usually contains both \(x\) and \(y\), so substitute both coordinates of the point.
- Horizontal tangent: numerator of \(\frac{dy}{dx}\) is zero. Vertical tangent: denominator is zero. Either way, solve together with the curve's equation.
- \(\frac{d}{dx}(a^x)=a^x\ln a\), from \(\ln y = x\ln a\).
- "Rate of change" means \(\frac{d}{dt}\). Link rates with the chain rule, e.g. \(\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}\). "Proportional to" becomes "\(=k\times\)", with a minus sign for a decrease.
Contents
14 sections
- What does it mean for a curve to be defined implicitly?
- How do you differentiate a term containing y?
- How do you find a tangent or normal to an implicit curve?
- How do you find stationary points and vertical tangents?
- How do you differentiate a to the power x?
- How do you turn a description into a differential equation?
- How does the chain rule connect two rates of change?
- How do the pieces fit in a longer question?
- Exam tips
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What does it mean for a curve to be defined implicitly?
An explicit equation gives \(y\) directly in terms of \(x\): \(y=x^3-4x\) or \(y=e^{2x}\sin x\). You differentiate these with the P3 rules. An implicit equation mixes \(x\) and \(y\) without \(y\) being the subject: \(x^2+y^2=25\), \(x^2y+y^3=10\), \(xe^y=2x+y\).
Rearranging is often impossible (try making \(y\) the subject of \(xe^y=2x+y\)), or it gives two branches, such as \(y=\pm\sqrt{25-x^2}\). 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 \(x^2+y^2=25\) 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 \(2x+2y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{x}{y}\): that is \(-\frac34\) at \((3,4)\) and \(+\frac34\) at \((3,-4)\).
How do you differentiate a term containing y?
Differentiate it as if the letter were \(x\), then multiply by \(\frac{dy}{dx}\). 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 \(x^2y^3\) 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.
| Term | Rule used | Derivative with respect to \(x\) |
|---|---|---|
| \(y^3\) | chain | \(3y^2\frac{dy}{dx}\) |
| \(\frac1y=y^{-1}\) | chain | \(-y^{-2}\frac{dy}{dx}\) |
| \(e^{2y}\) | chain | \(2e^{2y}\frac{dy}{dx}\) |
| \(\ln y\) | chain | \(\frac1y\frac{dy}{dx}\) |
| \(\sin y\) | chain | \(\cos y\frac{dy}{dx}\) |
| \(xy\) | product | \(y+x\frac{dy}{dx}\) |
| \(x^2y^3\) | product + chain | \(2xy^3+3x^2y^2\frac{dy}{dx}\) |
The method is always the same four steps: (1) differentiate every term on both sides; (2) collect the terms containing \(\frac{dy}{dx}\) on one side and everything else on the other; (3) factorise out \(\frac{dy}{dx}\); (4) divide.
The circle \(C\) has equation \(x^2+y^2-6x+4y-12=0\). (a) Find \(\frac{dy}{dx}\) 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.
(a) Differentiate term by term: \(2x+2y\frac{dy}{dx}-6+4\frac{dy}{dx}=0\). Collect and factorise: \((2y+4)\frac{dy}{dx}=6-2x\), so \(\dfrac{dy}{dx}=\dfrac{3-x}{y+2}\).
(b) Check \(P\) is on \(C\): \(49+1-42+4-12=0\) ✓. Gradient at \(P\): \(\frac{3-7}{1+2}=-\frac43\). Tangent: \(y-1=-\frac43(x-7)\), 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 \(\frac34\), and \(\frac34\times\left(-\frac43\right)=-1\), so the tangent is perpendicular to the radius, as it must be.
(a) The curve \(C_1\) has equation \(x^3-2xy+y^2=5\). Find the gradient of \(C_1\) at the point \((2,3)\). (b) The curve \(C_2\) has equation \(xe^y=2x+y\). Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\), and its value at the origin.
(a) Product rule on \(2xy\) (in a bracket because of the minus sign), chain rule on \(y^2\): \(3x^2-\left(2y+2x\frac{dy}{dx}\right)+2y\frac{dy}{dx}=0\). Collect: \((2y-2x)\frac{dy}{dx}=2y-3x^2\), so \(\frac{dy}{dx}=\frac{2y-3x^2}{2y-2x}\). The point is on \(C_1\) since \(8-12+9=5\). At \((2,3)\): \(\frac{6-12}{6-4}=\) \(-3\).
(b) Product rule on \(xe^y\); the right-hand side gives \(2+\frac{dy}{dx}\): \(e^y+xe^y\frac{dy}{dx}=2+\frac{dy}{dx}\). So \((xe^y-1)\frac{dy}{dx}=2-e^y\) and \(\dfrac{dy}{dx}=\dfrac{2-e^y}{xe^y-1}\). At \((0,0)\): \(\frac{2-1}{0-1}=\) \(-1\).
Differentiating \(2xy\) as \(2y\) (treating \(y\) as a constant) or as \(2\frac{dy}{dx}\) (forgetting the \(y\) term). It is a product: \(2y+2x\frac{dy}{dx}\). Watch the sign too: \(-2xy\) gives \(-2y-2x\frac{dy}{dx}\), so put the derivative in a bracket first.
How do you find a tangent or normal to an implicit curve?
The method is the one from P1: find the gradient \(m\) at the point, use \(-\frac1m\) for the normal, then \(y-y_1=m(x-x_1)\). The only change is that \(\frac{dy}{dx}\) 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.
The curve \(C\) has equation \(x^2y+y^3=10\). The point \(P(3,1)\) lies on \(C\). (a) Find \(\frac{dy}{dx}\) 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 \(P\).
(a) Product rule on \(x^2y\), chain rule on \(y^3\): \(2xy+x^2\frac{dy}{dx}+3y^2\frac{dy}{dx}=0\), so \(\dfrac{dy}{dx}=\dfrac{-2xy}{x^2+3y^2}\).
(b) At \(P(3,1)\): \(\frac{dy}{dx}=\frac{-6}{9+3}=-\frac12\). Then \(y-1=-\frac12(x-3)\), so \(x+2y-5=0\).
(c) Substitute \(x=5-2y\) into \(C\): \((5-2y)^2y+y^3=10\), which expands to \(5y^3-20y^2+25y-10=0\), i.e. \(y^3-4y^2+5y-2=0\). 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: \(1\times2+8=10\) ✓.
(d) Normal gradient \(=-1\div\left(-\frac12\right)=2\). So \(y-1=2(x-3)\), i.e. \(y=2x-5\).
How do you find stationary points and vertical tangents?
A stationary point has \(\frac{dy}{dx}=0\): 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 \(\frac{dx}{dy}=0\). Since \(\frac{dx}{dy}=1\big/\frac{dy}{dx}\) (P3), this happens where the denominator of \(\frac{dy}{dx}\) 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 \(y\).
The curve \(C\) has equation \(x^2+xy+y^2=12\). (a) Show that \(\frac{dy}{dx}=-\frac{2x+y}{x+2y}\). (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.
(a) Product rule on \(xy\): \(2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\), so \((x+2y)\frac{dy}{dx}=-(2x+y)\) and \(\frac{dy}{dx}=-\frac{2x+y}{x+2y}\) as required.
(b) Numerator zero: \(2x+y=0\), so \(y=-2x\). Substitute: \(x^2-2x^2+4x^2=12\), so \(3x^2=12\) and \(x=\pm2\). The points are \((2,-4)\) and \((-2,4)\). Check the denominator: at \((2,-4)\), \(x+2y=-6\ne0\).
(c) Denominator zero: \(x=-2y\). Substitute: \(4y^2-2y^2+y^2=12\), so \(y=\pm2\). 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.
Stopping at "\(y=-2x\)" or solving \(2x+y=0\) on its own. That line is where stationary points could be; the points themselves are where it meets the curve, so you must substitute back into the curve's equation.
How do you differentiate a to the power x?
The power rule does not apply to \(3^x\): 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 \(\frac{d}{dx}(e^x)=e^x\). For \(0<a<1\), \(\ln a<0\) and the gradient is negative everywhere. Notice that the gradient \(a^x\ln a\) is a constant multiple of the height \(a^x\): every exponential curve grows at a rate proportional to its own value, which is exactly why exponentials model growth.
(a) By taking natural logarithms, show that if \(y=a^x\), where \(a\) is a positive constant, then \(\frac{dy}{dx}=a^x\ln a\). (b) The tangent to \(y=3^x\) at the point \((2,9)\) meets the \(x\)-axis at \(A\). Find the exact \(x\)-coordinate of \(A\). (c) Differentiate \(y=4^{3x}\).
(a) \(\ln y=x\ln a\), so \(\frac1y\frac{dy}{dx}=\ln a\), and \(\frac{dy}{dx}=y\ln a=a^x\ln a\) as required.
(b) Gradient at \(x=2\): \(3^2\ln3=9\ln3\ (\approx9.89)\). Tangent: \(y-9=9\ln3\,(x-2)\). Put \(y=0\): \(-9=9\ln3\,(x-2)\), so \(x=2-\dfrac1{\ln3}\ (\approx1.09)\).
(c) \(\ln y=3x\ln4\), so \(\frac1y\frac{dy}{dx}=3\ln4\) and \(\dfrac{dy}{dx}=3\ln4\cdot4^{3x}\ (=6\ln2\cdot4^{3x})\).
How do you turn a description into a differential equation?
A differential equation is an equation containing a derivative. In context, "the rate of change of \(N\)" (with time) is \(\frac{dN}{dt}\). 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 \(A=\frac kB\).
- Take \(k>0\) and put the sign in yourself: a quantity that decreases gets a minus sign, \(\frac{dm}{dt}=-km\).
- If a rate is given at one moment ("when \(\theta=78\) 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 \(\frac{dV}{dt}=20-kh\).
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 \(\theta\) °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 \(\theta=78\) it is cooling at 2.4 °C per minute. Find the constant, and the rate of cooling when \(\theta=48\).
(a) \(\dfrac{dN}{dt}=kN\), where \(k>0\) is a constant.
(b) Decreasing, so a minus sign: \(\dfrac{dh}{dt}=-k\sqrt h\), \(k>0\).
(c) The difference is \(\theta-18\) and the temperature is falling: \(\frac{d\theta}{dt}=-k(\theta-18)\). Substitute: \(-2.4=-k(78-18)\), so \(k=0.04\). At \(\theta=48\): \(\frac{d\theta}{dt}=-0.04\times30=-1.2\), 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 \(\theta\) as functions of \(t\), is done on the P4 differential equations page.
Always define your variables and say "\(k\) is a positive constant". If the question uses a different letter or tells you the sign, use exactly that form; a "show that" with a given equation tells you which way round the constant goes.
How does the chain rule connect two rates of change?
Often you know one rate (how fast air is pumped in, \(\frac{dV}{dt}\)) and want another (how fast the radius grows, \(\frac{dr}{dt}\)). A geometric formula links the quantities, \(V=\frac43\pi r^3\), so you can find \(\frac{dV}{dr}\). 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 \(A=\pi r^2\), so \(\frac{dA}{dt}=2\pi r\times3=6\pi r\). When \(r=15\) cm the area is growing at \(90\pi\approx283\) cm² s⁻¹. The radius grows at a constant rate, but the area grows faster and faster.
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 \(\frac43\pi r^3\) and its surface area is \(4\pi r^2\).] 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.
(a) \(V=\frac43\pi r^3\), so \(\frac{dV}{dr}=4\pi r^2=400\pi\) when \(r=10\). Chain rule: \(\frac{dV}{dt}=\frac{dV}{dr}\times\frac{dr}{dt}\), so \(50=400\pi\times\frac{dr}{dt}\) and \(\dfrac{dr}{dt}=\dfrac1{8\pi}\approx0.0398\) cm s⁻¹.
(b) \(S=4\pi r^2\), so \(\frac{dS}{dr}=8\pi r=80\pi\). Then \(\frac{dS}{dt}=80\pi\times\frac1{8\pi}=\) 10 cm² s⁻¹.
Keep \(\frac{dr}{dt}\) exact for part (b): the \(\pi\) cancels and the answer is exact.
Writing the chain upside down, e.g. \(\frac{dr}{dt}=\frac{dV}{dt}\times\frac{dV}{dr}\). That would give \(50\times400\pi\), a radius growing at over 60 000 cm per second. Check that the fractions "cancel" to the rate you want, and that the size of the answer is sensible.
How do the pieces fit in a longer question?
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.
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 \(\frac13\pi r^2h\).] (a) Show that \(V=\frac{\pi h^3}{12}\). (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 \(\frac{dh}{dt}=\frac{4(20-kh)}{\pi h^2}\). (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).
(a) Similar triangles: \(\frac rh=\frac6{12}\), so \(r=\frac h2\). Then \(V=\frac13\pi\left(\frac h2\right)^2h=\frac{\pi h^3}{12}\) as required.
(b) \(\frac{dV}{dh}=\frac{\pi h^2}4=16\pi\) at \(h=8\). So \(\frac{dh}{dt}=\frac{dV}{dt}\div\frac{dV}{dh}=\frac{20}{16\pi}=\) \(\dfrac5{4\pi}\approx0.398\) cm s⁻¹.
(c) Net rate in = rate in − rate out: \(\frac{dV}{dt}=20-kh\). Also \(\frac{dV}{dt}=\frac{\pi h^2}4\frac{dh}{dt}\). Equate: \(\frac{\pi h^2}4\frac{dh}{dt}=20-kh\), so \(\frac{dh}{dt}=\frac{4(20-kh)}{\pi h^2}\) as required.
(d) Depth constant means \(\frac{dh}{dt}=0\): \(20-10k=0\), so \(k=2\).
(e) With \(k=2\), \(h=8\): \(\frac{dh}{dt}=\frac{4(20-16)}{64\pi}=\) \(\dfrac1{4\pi}\approx0.0796\) 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.
Exam tips for P4.4
Write the whole equation after differentiating, before rearranging. That line, with \(\frac{dy}{dx}\) attached to every \(y\) term and the product rule applied to every mixed term, is where the method is shown; the rearrangement is a short step after it.
Differentiate both sides. A number such as the 12 in \(x^2+xy+y^2=12\) becomes 0; a term such as \(2x+y\) on the right becomes \(2+\frac{dy}{dx}\). Then collect.
Set the numerator to zero to get a link like \(y=-2x\), substitute it into the curve, and give both coordinates of every point. Check that the denominator is non-zero at your points.
Start with a line such as \(\frac{dh}{dt}=\frac{dh}{dV}\times\frac{dV}{dt}\), then substitute. Give units (cm s⁻¹, cm² s⁻¹) and 3 significant figures unless an exact answer is asked for.
"Decreases at a rate proportional to …" needs a minus sign with \(k>0\). When a rate is quoted at one instant, substitute the values at that instant to find \(k\), and check the sign of your \(k\) makes sense.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \(\frac{d}{dx}(y^2)=2y\). | \(y\) depends on \(x\); differentiating with respect to \(x\) needs the chain rule. | \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\). |
| \(\frac{d}{dx}(xy)=\frac{dy}{dx}\). | \(xy\) is a product of two functions of \(x\). | Product rule: \(y+x\frac{dy}{dx}\). |
| The gradient at a point only needs its \(x\)-coordinate. | Implicit derivatives usually contain \(y\) as well, and one \(x\) can give several points. | Substitute both coordinates; find \(y\) from the curve first if needed. |
| Setting the numerator of \(\frac{dy}{dx}\) to zero gives the stationary points. | It gives a relationship between \(x\) and \(y\), a whole line of possible points. | Solve it together with the curve's equation. |
| \(\frac{d}{dx}(2^x)=x\cdot2^{x-1}\). | The power rule needs a constant power; here the power is the variable. | \(\frac{d}{dx}(2^x)=2^x\ln2\). |
| In connected rates, substitute the given radius into the formula before differentiating. | The formula then becomes a constant, whose derivative is 0. | Differentiate the general formula, then substitute the value at that instant. |
How this topic connects
P3 differentiation
Every step here is the chain and product rules from P3; \(\frac{dy}{dx}=1\big/\frac{dx}{dy}\) explains why a zero denominator means a vertical tangent, and \(a^x=e^{x\ln a}\) gives a second route to the derivative of \(a^x\).
P1 and P2 coordinate geometry
Tangents and normals use \(y-y_1=m(x-x_1)\) and perpendicular gradients from P1; circles from P2 give a geometric check (tangent perpendicular to radius).
Other P4 pages
P4.2 Parametric equations give \(\frac{dy}{dx}\) 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 \(\frac{dN}{dt}=-\lambda N\) (radioactive decay) and Newton's law of cooling are the physics versions of forming differential equations.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\frac{d}{dx}[g(y)]=g'(y)\frac{dy}{dx}\), e.g. \(\frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}\) | differentiating any term in \(y\) implicitly | No — learn it |
| \(\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\) | mixed terms such as \(xy\), \(x^2y^3\), \(xe^y\) | Yes |
| \(\frac{dy}{dx}=0\) (numerator \(=0\)); \(\frac{dx}{dy}=0\) (denominator \(=0\)) | horizontal and vertical tangents on an implicit curve | No — learn it |
| \(a^x=e^{x\ln a}\) | rewriting an exponential with base \(a\) | Yes |
| \(\frac{d}{dx}(a^x)=a^x\ln a\) | gradient of \(y=a^x\) | No — derive it |
| \(\frac{dA}{dt}=\frac{dA}{dr}\times\frac{dr}{dt}\) | connected rates of change | No — learn it |
| "\(A\) decreases at a rate proportional to \(B\)": \(\frac{dA}{dt}=-kB\), \(k>0\) | forming a differential equation | No — learn it |
| Surface area of a sphere \(4\pi r^2\) | rates involving spheres | Yes |
| Area of a circle \(\pi r^2\); volume of a sphere \(\frac43\pi r^3\); volume of a cone \(\frac13\pi r^2h\) | linking quantities in rates questions | No — learn it |
Check your understanding
1. Find \(\frac{dy}{dx}\) for the curve \(y^2=4x\).
\(2y\frac{dy}{dx}=4\), so \(\frac{dy}{dx}=\frac2y\). For example, at \((1,2)\) the gradient is 1 and at \((1,-2)\) it is \(-1\).
2. Find the gradient of the curve \(xy=6\) at the point \((2,3)\).
\(y+x\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac yx=-\frac32\).
3. What is the gradient of \(y=2^x\) where it crosses the \(y\)-axis?
\(\frac{dy}{dx}=2^x\ln2\), which is \(\ln2\approx0.693\) at \(x=0\).
4. The radius of a circle increases at \(\frac12\) cm s⁻¹. How fast is the area increasing when the radius is 5 cm?
\(\frac{dA}{dt}=2\pi r\times\frac{dr}{dt}=2\pi\times5\times\frac12=5\pi\approx15.7\) cm² s⁻¹.
5. Write a differential equation for: "the mass \(m\) of a melting block of ice decreases at a rate inversely proportional to \(m\)".
\(\frac{dm}{dt}=-\frac km\), where \(k\) is a positive constant. "Inversely proportional" puts \(m\) in the denominator; "decreases" gives the minus sign.
6. In worked example 4, show that the point \((-2,4)\) is on the curve and confirm the tangent there is horizontal.
\(4+(-2)(4)+16=12\) ✓. The numerator \(2x+y=-4+4=0\) and the denominator \(x+2y=-2+8=6\ne0\), so \(\frac{dy}{dx}=0\): the tangent is horizontal.
Specification checklist
- (P4 5.1) I can differentiate an equation in \(x\) and \(y\) implicitly, using the chain rule on terms in \(y\) and the product rule on mixed terms.
- (P4 5.1) I can find the gradient, tangent and normal at a point on an implicitly defined curve, and where a line meets the curve again.
- (P4 5.1) I can find stationary points, and points with vertical tangents, on an implicitly defined curve.
- (P4 5.1) I can differentiate \(a^x\) by taking logarithms, and use the result for tangents and gradients.
- (P4 5.2) I can form a simple differential equation from a description involving rates and proportionality, and find the constant from given information.
- (P4 5.2) I can use the chain rule to connect rates of change, such as volume, radius, depth and area with respect to time.