Edexcel past paper questions - KUMAR'S MATHS … Implicit Differentiation Page 1 Edexcel past paper questions ... (C4 June 2015, Q2) Author: vasanthakumaran.kand

C4 Implicit Differentiation Page 1

Edexcel past paper questions

Core Mathematics 4

Implicit Differentiation

Edited by: K V Kumaran

Email: [email protected]


Implicit Differentiation

1. Introduction

Normally, when differentiating, it is dealing with ‘explicit functions’ of x, where a

value of y is defined only in terms of x. However, some functions cannot be

rearranged into this form, and we cannot express y solely in terms of x, therefore,

we say y is given implicitly by x. Even so, given a value of x, a value for y can still

be found, after a bit of work.

e.g.

y = 2x2 − 3x + 4 is expressed explicitly in terms of x.

x2 + y2 − 6x + 2y = 0 is expressed implicitly.

In general, to differentiate an implicit function to find, we differentiate both sides

of the equation with respect to x. This allows you differentiate without making y

the subject first.

In order to differentiate both sides of the equation, you will end up having to

differentiate a term in y with respect to x. To do this, you follow the rule

𝑑

𝑑𝑥(𝑦𝑛) = 𝑛𝑦𝑛−1

𝑑𝑦

𝑑𝑥

Basically this means you differentiate y as it were x, then add on 𝑑𝑦

𝑑𝑥 onto the end

of it. Effectively this is what you do when differentiating explicitly, but the y

differentiates to 1, leaving only dy/dx.


Another problem is differentiating something like 4xy2. This can be implicitly

differentiated using the product rule, taking u = 4x and v = y2. Remembering to

leave dy/dx after each y term that has been differentiated, it becomes

𝑑

𝑑𝑥[𝑓(𝑥)𝑔(𝑦)] = 𝑓(𝑥)𝑔′(𝑦)

𝑑𝑦

𝑑𝑥 + 𝑓′(𝑥)𝑔(𝑦)

After differentiating each term in turn, you then need to rearrange the equation

to find 𝑑𝑦

𝑑𝑥 on its own.

nb: the function y=ax differentiates to axln(a), and this can be shown through implicit differentiation by taking logs of both sides.







Further examples,

Example 1

A curve is described by the equation

2x2 + 4y2 – 4x + 8xy + 26 = 0

Find an equation of the tangent to C at the point (3, -2), giving your answer in the

form ax + by + c = 0, where a, b and c are integers.

2x2 + 4y2 – 4x + 8xy + 26 = 0

4x + 8ydy

dx - 4 + 8y + 8x

dy

dx = 0 (a lot of students leave in the 26)

Rearrange to make dy

dx the subject

8ydy

dx + 8x

dy

dx = 4 – 8y -4x

dy

dx =

4 - 8y -4x 1 2y x

8y+8x 2y 2x

We finally have the gradient function so by substituting the values of x = 3 and

y = -2

Grad = 1

Using y = mx + c

-2 = 3 + c c = -5

Therefore the equation of the tangent is:

y = x – 5

The same principles are used in the following example to find turning points.


Example 2

A curve has equation x2 + 4xy – 3y2 + 28 = 0,

Find the coordinates of the turning points on the curve.

Using the ideas of implicit differentiation outlined in example (1) and taking care

with the product 4xy.

2x + 4xdy

dx + 4y – 6y

dy

dx = 0

(4x – 6y) dy

dx = -2x – 4y

dy x 2y

dx 2x 3y

The turning points occur when dy

dx=0.

Therefore x = -2y.

(Why have we not considered the denominator to be zero?)

Substituting into the original curve gives:

x2 + 4xy – 3y2 + 28 = 0

4y2 – 8y2 – 3y2 + 28 = 0

7y2 = 28 y = 2

Coordinates of turning points (-4,2) and (4,-2)


C4 Implicit differentiation past paper questions

1. A curve has equation

x2 + 2xy – 3y2 + 16 = 0.

Find the coordinates of the points on the curve where x

y

d

d = 0. (7)

(C4 June 2005, Q2)

2. A curve C is described by the equation

3x2 + 4y2 – 2x + 6xy – 5 = 0.

Find an equation of the tangent to C at the point (1, –2), giving your answer in the

form ax + by + c = 0, where a, b and c are integers. (7)

(C4 Jan 2006, Q1)

3. A curve C is described by the equation

3x2 – 2y2 + 2x – 3y + 5 = 0.

Find an equation of the normal to C at the point (0, 1), giving your answer in the

form ax + by + c = 0, where a, b and c are integers. (7)

(C4 June 2006, Q1)

4. A set of curves is given by the equation sin x + cos y = 0.5.

(a) Use implicit differentiation to find an expression for x

y

d

d. (2)

For – < x < and – < y < ,

(b) find the coordinates of the points where x

y

d

d = 0. (5)

(C4 Jan 2007, Q5)

5. A curve is described by the equation

x3 4y2 = 12xy.


(a) Find the coordinates of the two points on the curve where x = –8. (3)

(b) Find the gradient of the curve at each of these points. (6)

(C4 Jan 2008, Q5)

6. A curve has equation 3x2 – y2 + xy = 4. The points P and Q lie on the curve. The

gradient of the tangent to the curve is 38 at P and at Q.

(a) Use implicit differentiation to show that y – 2x = 0 at P and at Q. (6)

(b) Find the coordinates of P and Q. (3)

(C4 June 2008, Q4)

7. A curve C has the equation y2 – 3y = x3 + 8.

(a) Find x

y

d

d in terms of x and y. (4)

(b) Hence find the gradient of C at the point where y = 3. (3)

(C4 Jan 2009, Q1)

8. The curve C has the equation ye–2x = 2x + y2.

(a) Find x

y

d


The point P on C has coordinates (0, 1).

(b) Find the equation of the normal to C at P, giving your answer in the form ax +

by + c = 0, where a, b and c are integers. (4)

(C4 June 2009, Q4)

9. A curve C has equation

2x + y2 = 2xy.

Find the exact value of x

y

d

d at the point on C with coordinates (3, 2). (7)


(C4 June 2010, Q3)

10. Find the gradient of the curve with equation

ln y = 2x ln x, x > 0, y > 0,

at the point on the curve where x = 2. Give your answer as an exact value.

(7)

(C4 June 2011, Q5)

11. The curve C has the equation 2x + 3y2 + 3x2 y = 4x2.

The point P on the curve has coordinates (–1, 1).

(a) Find the gradient of the curve at P. (5)

(b) Hence find the equation of the normal to C at P, giving your answer in the form

ax + by + c = 0, where a, b and c are integers. (3)

(C4 June 2012, Q1)

12. A curve is described by the equation

x2 + 4xy + y2 + 27 = 0

(a) Find d

d

y

x in terms of x and y.

(5) A point Q lies on the curve. The tangent to the curve at Q is parallel to the y-axis. Given that the x-coordinate of Q is negative, (b) use your answer to part (a) to find the coordinates of Q.

(7)

(C4 June 2013, Q7)


13. The curve C has equation

3x–1 + xy –y2 +5 = 0

Show that d

d

y

x at the point (1, 3) on the curve C can be written in the form 31

ln( e )

, where λ

and μ are integers to be found. (7)

(C4 June 2013_R, Q2)

14. A curve C has the equation

x3 + 2xy – x – y3 – 20 = 0

(a) Find d

d

y

x in terms of x and y. (5)

(b) Find an equation of the tangent to C at the point (3, –2), giving your answer in the form ax + by + c = 0, where a, b and c are integers. (2)

(C4 June 2014, Q1)

15. x2 + y2 + 10x + 2y – 4xy = 10

(a) Find d

d

y

x in terms of x and y, fully simplifying your answer. (5)

(b) Find the values of y for which d

0d

y

x . (5)

(C4 June 2014_R, Q3)

16. The curve C has equation

x2 – 3xy – 4y2 + 64 = 0.

(a) Find x

y

d


(b) Find the coordinates of the points on C where x

y

d

d = 0.

(Solutions based entirely on graphical or numerical methods are not acceptable.) (6)

(C4 June 2015, Q2)

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Edexcel past paper questions - KUMAR'S MATHS … Implicit Differentiation Page 1 Edexcel past paper questions ... (C4 June 2015, Q2) Author: vasanthakumaran.kand