Mathematics guide, Unit 3, step 14 of 33Previous step Next step

Step 14: Implicit Differentiation and Inverse Functions

By the EconTutorials team. 12-minute read with 6 practice questions.

SyllabusAP 3.2–3.6 · 9709 P2/P3 2.4, 3.4 · IB HL 5.14–5.15

In 30 seconds

When $ x $ and $ y $ are tied together in one equation, such as $ x^2 + 3xy-y^2 = 3 $, differentiate both sides with respect to $ x $. Every term in $ y $ picks up a factor $ \dfrac{dy}{dx} $ by the chain rule; then collect and solve.
For a parametric curve $ x = f(t) $, $ y = g(t) $, the gradient is $ \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} $. You never need to eliminate $ t $.
For an inverse function, if $ f(a) = b $ then $ \left(f^{-1}\right)'(b) = \dfrac{1}{f'(a)} $. This gives $ \dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}} $, $ \dfrac{d}{dx}\cos^{-1}x = -\dfrac{1}{\sqrt{1-x^2}} $ and $ \dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1 + x^2} $.
Exam point: an implicit gradient usually contains both $ x $ and $ y $, so you need both coordinates of the point. And $ \left(f^{-1}\right)'(b) $ uses $ f’ $ at $ a $, the matching $ x $-value, never at $ b $.

Video lesson coming soonA short animated lesson for this step will appear here.

Curves that are not y = f(x)

Every rule so far has assumed that $ y $ is given as a formula in $ x $. Plenty of curves are not like that. The equation $ x^2 + 3xy-y^2 = 3 $ describes a perfectly good curve (Figure 1), but for $ x = 1 $ it has two points, $ (1, 1) $ and $ (1, 2) $, so it is not the graph of one function. You could solve the quadratic in $ y $ and differentiate each branch, but that is slow, and for many curves it is impossible.

The way round it uses nothing new. Along the curve, $ y $ still depends on $ x $, even if we cannot write the formula down. So differentiate both sides of the equation with respect to $ x $, treating $ y $ as an unknown function of $ x $. A term such as $ y^2 $ is then a composite (a function of $ y $, which is a function of $ x $), and the chain rule from Step 13 attaches a factor $ \dfrac{dy}{dx} $:

\[ \begin{gathered} \frac{d}{dx}\left(y^2\right) = 2y\frac{dy}{dx}, \qquad \frac{d}{dx}\left(\sin y\right) = \cos y\,\frac{dy}{dx}, \\ \frac{d}{dx}\left(e^{y}\right) = e^{y}\frac{dy}{dx}. \end{gathered} \]

Key termImplicit differentiation: differentiating every term of an equation in $ x $ and $ y $ with respect to $ x $, treating $ y $ as a function of $ x $. Each term in $ y $ gains a factor $ \dfrac{dy}{dx} $, and a mixed term such as $ xy $ needs the product rule: $ \dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx} $.

After differentiating, the job is pure algebra: gather the $ \dfrac{dy}{dx} $ terms on one side, factorise, and divide.

Worked example

The curve $ x^2 + 3xy-y^2 = 3 $ passes through $ (1, 2) $. Find $ \dfrac{dy}{dx} $, then the equations of the tangent and the normal at $ (1, 2) $.

Step 1Check the point: $ 1 + 6-4 = 3 $. It is on the curve.

Step 2Differentiate term by term, with the product rule on $ 3xy $: $ 2x + 3y + 3x\dfrac{dy}{dx}-2y\dfrac{dy}{dx} = 0 $.

Step 3Collect: $ (3x-2y)\dfrac{dy}{dx} = -(2x + 3y) $, so $ \dfrac{dy}{dx} = \dfrac{2x + 3y}{2y-3x} $.

Step 4At $ (1, 2) $: $ \dfrac{dy}{dx} = \dfrac{2 + 6}{4-3} = 8 $. Tangent: $ y-2 = 8(x-1) $, so $ y = 8x-6 $.

Step 5Normal: gradient $ -\tfrac{1}{8} $, so $ y-2 = -\tfrac{1}{8}(x-1) $. Multiply by 8: $ x + 8y-17 = 0 $.

At the other point with $ x = 1 $, $ (1, 1) $, the same formula gives $ \dfrac{2 + 3}{2-3} = -5 $. One formula, two different gradients: that is why you always need both coordinates.

The curve x² + 3xy − y² = 3, with its tangent and normal at (1, 2)

x² + 3xy − y² = 3tangent: y = 8x − 6normal: x + 8y − 17 = 0

Figure 1. Two branches, and for x = 1 two points: (1, 2) and (1, 1). No single formula y = f(x) covers the whole curve, but implicit differentiation gives the gradient at any point you name. At (1, 2) it is 8.

Common mistakeLeaving out $ \dfrac{dy}{dx} $ on a $ y $ term, or putting it on only one half of a product. $ \dfrac{d}{dx}(y^3) $ is $ 3y^2\dfrac{dy}{dx} $, not $ 3y^2 $, and $ \dfrac{d}{dx}(3xy) $ is $ 3y + 3x\dfrac{dy}{dx} $. A constant on the right-hand side differentiates to 0, so it plays no part in the formula for $ \dfrac{dy}{dx} $.

Worked example

Find the gradient of the curve $ e^{y} + xy = e $ at the point $ (0, 1) $.

Step 1Check: $ e^{1} + 0 = e $. The point is on the curve.

Step 2Differentiate: $ e^{y}\dfrac{dy}{dx} + y + x\dfrac{dy}{dx} = 0 $, so $ \dfrac{dy}{dx} = -\dfrac{y}{e^{y} + x} $.

Step 3At $ (0, 1) $: $ \dfrac{dy}{dx} = -\dfrac{1}{e} \approx -0.368 $.

There is no way to write $ y $ as a formula in $ x $ here, yet the gradient took three lines.

An economics example: the marginal rate of substitution. An indifference curve joins all the bundles of two goods that give a consumer the same utility. With utility $ U = xy^2 $, the bundles worth $ U = 144 $ satisfy $ xy^2 = 144 $: an implicit curve. Differentiating, $ y^2 + 2xy\dfrac{dy}{dx} = 0 $, so $ \dfrac{dy}{dx} = -\dfrac{y}{2x} $. At $ (4, 6) $ the slope is $ -\tfrac{3}{4} $: the consumer would give up 0.75 units of Y for one more unit of X. At $ (9, 4) $ it is only $ -\tfrac{2}{9} $. Economists call $ -\dfrac{dy}{dx} $ the marginal rate of substitution, and its fall along the curve (Figure 2) is the familiar diminishing MRS. The same algebra on an isoquant gives a firm’s marginal rate of technical substitution.

Indifference curve xy² = 144

xy² = 144tangent at (4, 6): slope −3/4tangent at (9, 4): slope −2/9

Figure 2. Every bundle on the curve gives the same utility. The slope is steep at (4, 6), where the consumer has little X and would give up 0.75 of Y for one more unit of it, and shallow at (9, 4), where the MRS has fallen to 2/9.

Parametric curves

Sometimes a curve is traced out by a moving point, and each coordinate is given in terms of a third variable, the parameter, usually $ t $ or $ \theta $: $ x = f(t) $, $ y = g(t) $. To find the gradient, the chain rule says $ \dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt} $. Divide through and you have the rule.

\[ \frac{dy}{dx} = \frac{\;dy/dt\;}{dx/dt} \qquad \text{wherever } \frac{dx}{dt} \neq 0 \]

The tangent is horizontal where $ \dfrac{dy}{dt} = 0 $ (and $ \dfrac{dx}{dt} \neq 0 $), and vertical where $ \dfrac{dx}{dt} = 0 $ (and $ \dfrac{dy}{dt} \neq 0 $).

Worked example

A curve has parametric equations $ x = t^3 $, $ y = 3t-t^2 $. (a) Find the tangent at the point where $ t = 1 $. (b) Find the points where the tangent is horizontal or vertical.

Step 1$ \dfrac{dx}{dt} = 3t^2 $ and $ \dfrac{dy}{dt} = 3-2t $, so $ \dfrac{dy}{dx} = \dfrac{3-2t}{3t^2} $.

Step 2(a) At $ t = 1 $: the point is $ (1, 2) $ and the gradient is $ \tfrac{1}{3} $. Tangent: $ y-2 = \tfrac{1}{3}(x-1) $, that is $ x-3y + 5 = 0 $.

Step 3(b) Horizontal: $ 3-2t = 0 $, so $ t = \tfrac{3}{2} $, giving the point $ \left(\tfrac{27}{8}, \tfrac{9}{4}\right) $.

Step 4Vertical: $ 3t^2 = 0 $, so $ t = 0 $, giving $ (0, 0) $. There $ \dfrac{dy}{dt} = 3 \neq 0 $, so the tangent really is vertical.

Substituting $ t $ into $ x $, $ y $ and $ \dfrac{dy}{dx} $ separately, and writing all three down, avoids the classic slip of using $ t = 1 $ as if it were the x-coordinate.

For concavity you also need $ \dfrac{d^2y}{dx^2} $, and it is not $ \dfrac{d^2y/dt^2}{d^2x/dt^2} $. Apply the gradient rule again, to the function $ \dfrac{dy}{dx} $: $ \dfrac{d^2y}{dx^2} = \dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) \Big/ \dfrac{dx}{dt} $. For the curve above this gives $ -\tfrac{4}{9} $ at $ t = 1 $, so the curve is concave down there. This second-derivative form is AP BC only; Step 28 takes parametric curves further.

Exam tipImplicit differentiation is AP topic 3.2, Cambridge 9709 Papers 2 and 3 (topic 2.4, first derivatives only) and IB HL 5.14. Parametric gradients are on 9709 Papers 2 and 3 and in AP Unit 9, which is AP BC only. Cambridge questions nearly always end with a tangent or normal, often in the form $ ax + by + c = 0 $ with integer coefficients.

The derivative of an inverse function

If $ f $ is one-to-one, its inverse $ f^{-1} $ undoes it, and the graph of $ y = f^{-1}(x) $ is the reflection of $ y = f(x) $ in the line $ y = x $ (Step 3). Reflection swaps $ x $ and $ y $. The point $ (a, b) $ on $ f $ becomes $ (b, a) $ on $ f^{-1} $, and a tangent that rises $ m $ for every 1 across becomes one that rises 1 for every $ m $ across. So the gradient $ m $ turns into $ \dfrac{1}{m} $.

Key termDerivative of an inverse function: if $ f(a) = b $ and $ f'(a) \neq 0 $, then $ \left(f^{-1}\right)'(b) = \dfrac{1}{f'(a)} $. In Leibniz notation, $ \dfrac{dx}{dy} = 1 \Big/ \dfrac{dy}{dx} $.

The algebra agrees. Differentiate $ f\big(f^{-1}(x)\big) = x $ with the chain rule: $ f’\big(f^{-1}(x)\big) \cdot \left(f^{-1}\right)'(x) = 1 $. Put $ x = b $, so that $ f^{-1}(b) = a $, and divide by $ f'(a) $. If $ f'(a) = 0 $ the tangent to $ f $ is horizontal, its reflection is vertical, and $ f^{-1} $ is not differentiable at $ b $.

Worked example

Let $ f(x) = 2x + \cos x $. Explain why $ f $ has an inverse, and find $ \left(f^{-1}\right)'(1) $.

Step 1$ f'(x) = 2-\sin x \geq 1 $ for every $ x $, so $ f $ is always increasing and therefore one-to-one.

Step 2You cannot solve $ y = 2x + \cos x $ for $ x $, and you do not need to. Find $ a $ with $ f(a) = 1 $: try $ a = 0 $, since $ 0 + \cos 0 = 1 $.

Step 3$ f'(0) = 2-0 = 2 $, so $ \left(f^{-1}\right)'(1) = \dfrac{1}{2} $.

Figure 3 shows the two tangents: gradient 2 at $ (0, 1) $ on $ f $, and gradient $ \tfrac{1}{2} $ at $ (1, 0) $ on its inverse.

f(x) = 2x + cos x and its inverse, reflected in y = x

y = xy = f(x) = 2x + cos xy = f⁻¹(x)

Figure 3. The point (0, 1) on f reflects to (1, 0) on the inverse. The tangent with gradient 2 reflects to a tangent with gradient 1/2: rise and run swap places.

Common mistakeEvaluating $ f’ $ at the wrong point. In the example, $ \left(f^{-1}\right)'(1) $ is not $ \dfrac{1}{f'(1)} $. The 1 is an output of $ f $, a y-value. First go back to the input that produced it ($ a = 0 $), then take the reciprocal of $ f’ $ there. AP questions often give this as a table of values, and the trap is the same.

An economics example: elasticity from inverse demand. Economists usually write demand with price as a function of quantity, for instance $ P = 100e^{-0.02Q} $. Price elasticity needs $ \dfrac{dQ}{dP} $, the other way round, and the inverse rule supplies it: $ \dfrac{dQ}{dP} = 1 \Big/ \dfrac{dP}{dQ} $. Here $ \dfrac{dP}{dQ} = -2e^{-0.02Q} = -0.02P $, so

\[ \varepsilon = \frac{dQ}{dP}\cdot\frac{P}{Q} = \frac{1}{-0.02P}\cdot\frac{P}{Q} = -\frac{50}{Q}. \]

At $ Q = 50 $ (price about $36.79) the elasticity is exactly $ -1 $: demand is unit elastic, which is where revenue is at its largest. Below 50 units demand is elastic, and above it inelastic. We never had to solve for $ Q $ in terms of $ P $.

Inverse trigonometric functions

The functions $ \sin^{-1}x $, $ \cos^{-1}x $ and $ \tan^{-1}x $ (also written $ \arcsin x $, $ \arccos x $, $ \arctan x $) return an angle in radians, restricted to a principal range so that each has one output: $ \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] $ for $ \sin^{-1} $, $ [0, \pi] $ for $ \cos^{-1} $ and $ \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right) $ for $ \tan^{-1} $. Implicit differentiation finds each derivative.

Take $ y = \sin^{-1}x $. Then $ \sin y = x $. Differentiate with respect to $ x $: $ \cos y\,\dfrac{dy}{dx} = 1 $, so $ \dfrac{dy}{dx} = \dfrac{1}{\cos y} $. On the principal range $ \cos y \geq 0 $, so $ \cos y = +\sqrt{1-\sin^2 y} = \sqrt{1-x^2} $. For $ y = \tan^{-1}x $, $ \tan y = x $ gives $ \sec^2 y\,\dfrac{dy}{dx} = 1 $, and $ \sec^2 y = 1 + \tan^2 y = 1 + x^2 $. The same method on $ \cos y = x $ gives the third result.

Function	Derivative	Valid for	With the chain rule
$ \sin^{-1}x $	$ \dfrac{1}{\sqrt{1-x^2}} $	$ -1 < x < 1 $	$ \dfrac{u’}{\sqrt{1-u^2}} $
$ \cos^{-1}x $	$ -\dfrac{1}{\sqrt{1-x^2}} $	$ -1 < x < 1 $	$ -\dfrac{u’}{\sqrt{1-u^2}} $
$ \tan^{-1}x $	$ \dfrac{1}{1 + x^2} $	all $ x $	$ \dfrac{u’}{1 + u^2} $

The first two differ only in sign, because $ \sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2} $, a constant.

Worked example

Differentiate (a) $ \tan^{-1}3x $ (b) $ \sin^{-1}\left(x^2\right) $ (c) $ x\cos^{-1}x $.

Step 1(a) Inside $ u = 3x $, $ u’ = 3 $: $ \dfrac{3}{1 + 9x^2} $.

Step 2(b) Inside $ u = x^2 $, $ u’ = 2x $: $ \dfrac{2x}{\sqrt{1-x^4}} $.

Step 3(c) Product rule: $ 1 \cdot \cos^{-1}x + x \cdot \left(-\dfrac{1}{\sqrt{1-x^2}}\right) = \cos^{-1}x-\dfrac{x}{\sqrt{1-x^2}} $.

Each one is a standard derivative from the table combined with a rule you already know.

Exam tipAP (topic 3.4) and IB HL (5.15) use all three inverse trigonometric derivatives. Cambridge 9709 Paper 3 requires only $ \tan^{-1}x $; $ \sin^{-1}x $ and $ \cos^{-1}x $ are not on that syllabus. It is worth remembering $ \dfrac{d}{dx}\tan^{-1}\dfrac{x}{a} = \dfrac{a}{a^2 + x^2} $: read backwards, it becomes a standard integral later on.

Choosing a method, and higher derivatives

AP calls it “selecting procedures” (topic 3.5): look at the form of the expression and pick the rule before you write anything.

You are given	Method
$ y $ as a formula in $ x $	Chain, product and quotient rules (Steps 12 and 13)
One equation mixing $ x $ and $ y $	Implicit differentiation
$ x $ and $ y $ both in terms of $ t $	$ \dfrac{dy}{dt} \div \dfrac{dx}{dt} $
A value of $ \left(f^{-1}\right)’ $ with no formula for $ f^{-1} $	$ \dfrac{1}{f'(a)} $, where $ f(a) = b $
The variable in both the base and the power, as in $ x^{1/x} $	Take logarithms first, then differentiate implicitly

The last row is logarithmic differentiation. Neither the power rule nor the exponential rule fits $ y = x^{1/x} $ (for $ x > 0 $), but $ \ln y = \dfrac{\ln x}{x} $ is easy. Differentiating, $ \dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{1-\ln x}{x^2} $, so $ \dfrac{dy}{dx} = x^{1/x}\,\dfrac{1-\ln x}{x^2} $, which is zero only at $ x = e $. None of the three syllabuses names this method, but it is a tidy use of implicit differentiation.

Higher-order derivatives (AP topic 3.6) are just derivatives of derivatives: $ f”(x) $ or $ \dfrac{d^2y}{dx^2} $, then $ f”'(x) $, then $ f^{(4)}(x) $ and so on. For an implicit curve, differentiate the expression for $ \dfrac{dy}{dx} $ again, still treating $ y $ as a function of $ x $, substitute the first derivative, and tidy up with the curve’s own equation.

Worked example

For the curve $ x^2-y^2 = 9 $, show that $ \dfrac{d^2y}{dx^2} = -\dfrac{9}{y^3} $, and evaluate it at $ (5, 4) $.

Step 1First derivative: $ 2x-2y\dfrac{dy}{dx} = 0 $, so $ \dfrac{dy}{dx} = \dfrac{x}{y} $.

Step 2Quotient rule, keeping $ \dfrac{dy}{dx} $ wherever $ y $ is differentiated: $ \dfrac{d^2y}{dx^2} = \dfrac{y-x\dfrac{dy}{dx}}{y^2} = \dfrac{y-\dfrac{x^2}{y}}{y^2} = \dfrac{y^2-x^2}{y^3} $.

Step 3The curve says $ x^2-y^2 = 9 $, so $ y^2-x^2 = -9 $ and $ \dfrac{d^2y}{dx^2} = -\dfrac{9}{y^3} $.

Step 4At $ (5, 4) $: check $ 25-16 = 9 $. Then $ \dfrac{dy}{dx} = \tfrac{5}{4} $ and $ \dfrac{d^2y}{dx^2} = -\dfrac{9}{64} $.

The negative value says that the upper branch of this hyperbola is concave down at $ (5, 4) $.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. The curve $ x^2y + y = 10 $ passes through $ (2, 2) $. Find the gradient there. Show worked answer

Product rule on $ x^2y $: $ 2xy + x^2\dfrac{dy}{dx} + \dfrac{dy}{dx} = 0 $.
So $ \dfrac{dy}{dx} = -\dfrac{2xy}{x^2 + 1} $.
At $ (2, 2) $: $ -\dfrac{8}{5} $.

EasyQ2. Differentiate $ y = \cos^{-1}(5x) $. Show worked answer

Inside $ u = 5x $, $ u’ = 5 $.
$ \dfrac{dy}{dx} = -\dfrac{5}{\sqrt{1-25x^2}} $, for $ -\tfrac{1}{5} < x < \tfrac{1}{5} $.

MediumQ3. Let $ f(x) = x^5 + x^3 + 1 $, and let $ g $ be the inverse of $ f $. Find $ g'(3) $. Show worked answer

Find $ a $ with $ f(a) = 3 $: $ f(1) = 1 + 1 + 1 = 3 $, so $ a = 1 $.
$ f'(x) = 5x^4 + 3x^2 $, so $ f'(1) = 8 $.
$ g'(3) = \dfrac{1}{f'(1)} = \dfrac{1}{8} $.

MediumQ4. A curve has parametric equations $ x = 2\sin t $, $ y = \cos 2t $. Find the equation of the tangent at the point where $ t = \tfrac{\pi}{6} $. Show worked answer

$ \dfrac{dx}{dt} = 2\cos t $, $ \dfrac{dy}{dt} = -2\sin 2t $.
$ \dfrac{dy}{dx} = \dfrac{-2\sin 2t}{2\cos t} = \dfrac{-4\sin t\cos t}{2\cos t} = -2\sin t $.
At $ t = \tfrac{\pi}{6} $: point $ \left(1, \tfrac{1}{2}\right) $, gradient $ -1 $.
Tangent: $ y-\tfrac{1}{2} = -(x-1) $, so $ y = \tfrac{3}{2}-x $.
(Check: $ \cos 2t = 1-2\sin^2 t $, so the curve is the parabola $ y = 1-\tfrac{1}{2}x^2 $, whose gradient $ -x $ is $ -1 $ at $ x = 1 $.)

MediumQ5. The curve $ y^3 + y = x $ passes through $ (2, 1) $. Find $ \dfrac{dy}{dx} $ and $ \dfrac{d^2y}{dx^2} $ in terms of $ y $, and evaluate both at $ (2, 1) $. Show worked answer

$ 3y^2\dfrac{dy}{dx} + \dfrac{dy}{dx} = 1 $, so $ \dfrac{dy}{dx} = \dfrac{1}{3y^2 + 1} = (3y^2 + 1)^{-1} $.
Chain rule again: $ \dfrac{d^2y}{dx^2} = -(3y^2 + 1)^{-2} \cdot 6y\dfrac{dy}{dx} = -\dfrac{6y}{(3y^2 + 1)^3} $.
At $ y = 1 $: $ \dfrac{dy}{dx} = \dfrac{1}{4} $ and $ \dfrac{d^2y}{dx^2} = -\dfrac{6}{64} = -\dfrac{3}{32} $.

ExamQ6. A curve has equation $ x^2 + xy + 2y^2 = 8 $. (a) Show that $ \dfrac{dy}{dx} = -\dfrac{2x + y}{x + 4y} $. [3] (b) Find the equation of the normal to the curve at the point $ (2, 1) $, giving your answer in the form $ ax + by + c = 0 $. [3] (c) Find the exact coordinates of the points on the curve where the tangent is parallel to the x-axis. [4] Show worked answer

(a) $ 2x + \left(y + x\dfrac{dy}{dx}\right) + 4y\dfrac{dy}{dx} = 0 $. [M1 A1]
$ (x + 4y)\dfrac{dy}{dx} = -(2x + y) $, so $ \dfrac{dy}{dx} = -\dfrac{2x + y}{x + 4y} $. [A1]
(b) At $ (2, 1) $: $ \dfrac{dy}{dx} = -\dfrac{5}{6} $. [B1]
Normal gradient $ \tfrac{6}{5} $: $ y-1 = \tfrac{6}{5}(x-2) $. [M1]
$ 5y-5 = 6x-12 $, so $ 6x-5y-7 = 0 $. [A1]
(c) Parallel to the x-axis when $ 2x + y = 0 $, so $ y = -2x $. [M1]
Substitute: $ x^2-2x^2 + 8x^2 = 8 $, so $ 7x^2 = 8 $. [M1]
$ x = \pm\sqrt{\tfrac{8}{7}} = \pm\tfrac{2\sqrt{14}}{7} $. [A1]
Points $ \left(\tfrac{2\sqrt{14}}{7}, -\tfrac{4\sqrt{14}}{7}\right) $ and $ \left(-\tfrac{2\sqrt{14}}{7}, \tfrac{4\sqrt{14}}{7}\right) $; at both, $ x + 4y = -7x \neq 0 $, so the gradient is defined. [A1]
Total: 10 marks.

Check yourself

1. Differentiating implicitly with respect to $ x $, what is $ \dfrac{d}{dx}\left(y^2\right) $?

$ y $ is a function of $ x $, so $ y^2 $ is a composite. The chain rule gives $ 2y $ times the derivative of the inside, $ \dfrac{dy}{dx} $.

2. $ g $ is the inverse of $ f $, with $ f(3) = 7 $ and $ f'(3) = 5 $. What is $ g'(7) $?

$ f(3) = 7 $ means $ g(7) = 3 $, so $ g'(7) = \dfrac{1}{f'(3)} = \dfrac{1}{5} $.

3. What is the gradient of $ y = \tan^{-1}x $ at $ x = 1 $?

$ \dfrac{dy}{dx} = \dfrac{1}{1 + x^2} = \dfrac{1}{2} $ at $ x = 1 $. The value $ \tfrac{\pi}{4} $ is $ \tan^{-1}1 $ itself, the height of the graph, not its gradient.

Your score: 0 out of 3

The short version

Implicit differentiation: differentiate every term with respect to $ x $; each $ y $ term gains $ \dfrac{dy}{dx} $, and $ xy $ needs the product rule. Use both coordinates of the point.
Parametric curves: $ \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} $. Horizontal tangents where $ \dfrac{dy}{dt} = 0 $, vertical where $ \dfrac{dx}{dt} = 0 $.
Inverses: $ \left(f^{-1}\right)'(b) = \dfrac{1}{f'(a)} $ where $ f(a) = b $. Reflection in $ y = x $ turns gradient $ m $ into $ \dfrac{1}{m} $.
$ \dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}} $, $ \dfrac{d}{dx}\cos^{-1}x = -\dfrac{1}{\sqrt{1-x^2}} $, $ \dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1 + x^2} $ (9709 needs only $ \tan^{-1} $).
Implicit second derivatives: differentiate $ \dfrac{dy}{dx} $ again, substitute $ \dfrac{dy}{dx} $, then simplify with the curve’s equation.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

Function	Derivative	Valid for	With the chain rule
\( \sin^{-1}x \)	\( \dfrac{1}{\sqrt{1-x^2}} \)	\( -1 < x < 1 \)	\( \dfrac{u’}{\sqrt{1-u^2}} \)
\( \cos^{-1}x \)	\( -\dfrac{1}{\sqrt{1-x^2}} \)	\( -1 < x < 1 \)	\( -\dfrac{u’}{\sqrt{1-u^2}} \)
\( \tan^{-1}x \)	\( \dfrac{1}{1 + x^2} \)	all \( x \)	\( \dfrac{u’}{1 + u^2} \)

You are given	Method
\( y \) as a formula in \( x \)	Chain, product and quotient rules (Steps 12 and 13)
One equation mixing \( x \) and \( y \)	Implicit differentiation
\( x \) and \( y \) both in terms of \( t \)	\( \dfrac{dy}{dt} \div \dfrac{dx}{dt} \)
A value of \( \left(f^{-1}\right)’ \) with no formula for \( f^{-1} \)	\( \dfrac{1}{f'(a)} \), where \( f(a) = b \)
The variable in both the base and the power, as in \( x^{1/x} \)	Take logarithms first, then differentiate implicitly

Step 14: Implicit Differentiation and Inverse Functions

Curves that are not y = f(x)

Parametric curves

The derivative of an inverse function

Inverse trigonometric functions

Choosing a method, and higher derivatives

Practise: 6 questions

Check yourself

The short version

Further reading on EconTutorials

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