Pure: Differentiation Depth
Differentiation Depth consolidates the chain, product and quotient rules onto the harder settings A Level Maths actually examines them in: curves defined implicitly (where y is not given as a formula in x), curves defined parametrically (x and y both given in terms of a third variable t), connected rates of change (linking two changing quantities through a shared variable, usually time), and optimisation problems where a real quantity has to be written as a function of one variable before it can be differentiated. The examiner is testing whether a student can choose the right technique when the question does not name it, keep dy/dx notation and the chain rule correct when y appears inside another function of y, and justify a maximum or minimum rather than just finding where the gradient is zero.
Method
- Decide which technique applies before differentiating anything. If y is given explicitly in terms of x, differentiate directly with the chain, product or quotient rule. If x and y are both given in terms of a parameter t, use dy/dx = (dy/dt) / (dx/dt). If the equation mixes x and y and is not solved for y, differentiate implicitly.
- For implicit differentiation, differentiate every term of the equation with respect to x. Differentiate an x-only term as normal. Differentiate a y-only term (such as y^2 or y^3) using the chain rule, giving an extra factor of dy/dx (e.g. d/dx(y^2) = 2y dy/dx). Differentiate a mixed term such as xy with the product rule: d/dx(xy) = x dy/dx + y.
- After implicit differentiation, collect every term containing dy/dx on one side, factorise dy/dx out, and divide to make dy/dx the subject. To evaluate the gradient at a point, substitute BOTH the x-value and the y-value into the resulting expression, since it contains y as well as x.
- For connected rates of change, identify the two rates in the question and the variable that links them (commonly time, t). Use the chain rule to connect them, e.g. dV/dt = dV/dr x dr/dt, then substitute the known numerical rate and the value of the linking variable at the instant asked about.
- For parametric differentiation, find dx/dt and dy/dt separately and divide. For a parametric second derivative, do NOT differentiate dy/dx directly with respect to x; instead find d/dt(dy/dx) and divide by dx/dt.
- Learn the extra standard derivatives beyond sin, cos, e^x and ln(x): d/dx(tan x) = sec^2(x), d/dx(sec x) = sec(x)tan(x), d/dx(cosec x) = -cosec(x)cot(x), d/dx(cot x) = -cosec^2(x), and d/dx(a^x) = a^x ln(a), applying the chain rule whenever the argument is not simply x.
- For an optimisation problem, use any given constraint to write the quantity being optimised as a function of a single variable, differentiate, set the derivative to zero and solve. Justify a maximum or minimum using the second derivative (or a sign check either side), and reject any solution that falls outside the physically valid range given by the context.
Worked example
The curve C has equation x^2 + y^2 - 4x + 6y = 12. Find dy/dx in terms of x and y, and hence find the gradient of C at the point (5, 1).
- Differentiate every term of the equation with respect to x: d/dx(x^2) + d/dx(y^2) - d/dx(4x) + d/dx(6y) = d/dx(12), giving 2x + 2y(dy/dx) - 4 + 6(dy/dx) = 0.
- Collect the dy/dx terms on one side: (2y + 6)(dy/dx) = 4 - 2x.
- Divide to make dy/dx the subject: dy/dx = (4 - 2x) / (2y + 6), which simplifies to (2 - x) / (y + 3).
- Check the point (5, 1) lies on C: 5^2 + 1^2 - 4(5) + 6(1) = 25 + 1 - 20 + 6 = 12, confirmed.
- Substitute x = 5 and y = 1 into dy/dx: (2 - 5) / (1 + 3) = -3/4.
- Final answer: the gradient of C at (5, 1) is -3/4.
Practice questions
Try each question, then tap to reveal the answer.
Q1State the rule used to find dy/dx when x and y are both given in terms of a parameter t.Show answer
Answer: dy/dx = (dy/dt) / (dx/dt).
Q2A curve has parametric equations x = t^2 + 1, y = 3t - 2. Find dy/dx in terms of t.Show answer
Answer: dx/dt = 2t and dy/dt = 3, so dy/dx = 3/(2t).
Q3Find dy/dx for the curve x^2 y = 8, using implicit differentiation.Show answer
Answer: By the product rule, 2xy + x^2(dy/dx) = 0, so dy/dx = -2xy/x^2 = -2y/x.
Q4The radius of a circular oil spill is increasing at a constant rate of 0.4 m/s. Find the exact rate of increase of the area of the spill, in m^2/s, at the instant the radius is 5 m.Show answer
Answer: A = pi*r^2, so dA/dr = 2*pi*r. dA/dt = dA/dr x dr/dt = 2*pi*r x 0.4 = 0.8*pi*r. At r = 5, dA/dt = 4*pi m^2/s (about 12.6 m^2/s).
Q5Explain why, when differentiating y^3 implicitly with respect to x, the result is 3y^2(dy/dx) and not simply 3y^2.Show answer
Answer: y is itself a function of x, so the chain rule applies: d/dx(y^3) = d/dy(y^3) x dy/dx = 3y^2 x dy/dx. The extra factor of dy/dx comes from differentiating y with respect to x rather than treating y as the variable of differentiation.
Q6Differentiate y = tan(3x) with respect to x.Show answer
Answer: dy/dx = 3sec^2(3x), using the chain rule on the standard derivative of tan(x).
Q7A curve has parametric equations x = t^2, y = t^3 - 3t. Find dy/dx in terms of t, and find the coordinates of the two stationary points of the curve.Show answer
Answer: dx/dt = 2t and dy/dt = 3t^2 - 3, so dy/dx = 3(t^2-1)/(2t). Stationary points occur where dy/dt = 0 (and dx/dt is not 0): t = 1 gives (1, -2); t = -1 gives (1, 2).
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
The curve C has equation x^2 + xy + y^2 = 7. The point P(1, 2) lies on C. (a) Find dy/dx in terms of x and y. (3) (b) Find the equation of the tangent to C at P, giving your answer in the form ax + by = c, where a, b and c are integers. (3)
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Air is pumped into a spherical balloon at a constant rate of 15 cm^3/s. Find the rate of increase of the radius of the balloon, in cm/s, at the instant the radius is 10 cm. Give your answer to 3 significant figures.
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An open-top box is made by cutting a square of side x cm from each corner of a square sheet of card of side 20 cm, then folding up the sides. Let V cm^3 be the volume of the box. (a) Show that V = 4x^3 - 80x^2 + 400x. (2) (b) Find dV/dx, and find the value of x, in the domain 0 < x < 10, for which V is stationary. (3) (c) By finding d^2V/dx^2, show that this value of x gives a maximum volume, and find the exact maximum volume. (3)
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