A-Level Maths (Edexcel 9MA0) · Pure Year 2 · Chapter 9: Differentiation · Lesson 6

Implicit differentiation

113 marks

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Quick questions to start.

A1
Needed today
Differentiate $x^{2} e^{3 x}$ with respect to $x$ .
[2 marks]
Answer
A2
Needed today
A line has gradient $\frac{2}{5}$ . Write down the gradient of a line perpendicular to it, and find the equation of that perpendicular line through $(1, 3)$ .
[2 marks]
Answer
A3
Common slip
$y$ is a function of $x$ . Find $\frac{d}{d x} (y^{3})$ in terms of $y$ and $\frac{d y}{d x}$ .
[2 marks]
Answer
A4
Common slip
$y$ is a function of $x$ . Find $\frac{d}{d x} (4 x y)$ .
[2 marks]
Answer
A5
From chapter 9
Write $3^{x}$ in the form $e^{k x}$ , and hence differentiate $3^{x}$ .
[2 marks]
Answer

Example, then your turn

A-Level

Your teacher works through each example with you. Then try the one beside it.

Example 1

A curve has equation $x^{2} + 3 x y + y^{2} = 11$ . Find $\frac{d y}{d x}$ in terms of $x$ and $y$ .

Hint Differentiate every term with respect to

x

3 x y

needs the product rule, and

y^{2}

gives

2 y \frac{d y}{d x}

B1
A curve has equation $x^{3} + 2 x y - y^{2} = 7$ . Find $\frac{d y}{d x}$ in terms of $x$ and $y$ .
[4 marks]
Answer

Example 2

The curve $x^{2} + x y - y^{2} = 11$ passes through $(3, 2)$ . Find an equation of the normal to the curve at $(3, 2)$ . Give it in the form $a x + b y + c = 0$ .

Hint Differentiate, then put

x = 3

and

y = 2

in before you rearrange: it is quicker.

B2
The curve $2 x^{2} - x y + y^{2} = 16$ passes through $(2, 4)$ . Find an equation of the tangent to the curve at $(2, 4)$ . Give it in the form $a x + b y + c = 0$ , where $a$ , $b$ and $c$ are integers.
[5 marks]
Answer

Example 3

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

Hint

x e^{y}

is a product. The derivative of

e^{y}

with respect to

x

e^{y} \frac{d y}{d x}

B3
A curve has equation $e^{x y} + 3 y = 4$ . Find the gradient of the curve at the point $(0, 1)$ .
[4 marks]
Answer

Example 4

Find the coordinates of the stationary points on the curve $x^{2} - x y + y^{2} = 12$ .

Hint Stationary means

\frac{d y}{d x} = 0

, so the top of the fraction is zero. That gives a link between

x

and

y

: put it back into the curve.

B4
Find the coordinates of the stationary points on the curve $x^{2} + 2 x y + 3 y^{2} = 18$ .
[5 marks]
Answer

Example 5

Find the coordinates of the points on the curve $x^{2} + x y + y^{2} = 27$ where the tangent is vertical.

Hint A vertical tangent means the bottom of

\frac{d y}{d x}

is zero.

B5
Find the coordinates of the points on the curve $x^{2} - 4 x y + 5 y^{2} = 4$ where the tangent is vertical. Give exact answers.
[5 marks]
Answer

Practice

A-Level

C1
A curve has equation $x^{2} + 3 x y + y^{3} = 5$ . Ali differentiates and writes $2 x + 3 \frac{d y}{d x} + 3 y^{2} \frac{d y}{d x} = 0$ . Explain his mistake, and find the correct gradient at $(1, 1)$ .
[5 marks]
Answer

C2
A curve has equation $x^{2} + y^{2} + x y = 7$ . Beth differentiates the $7$ and gets $7$ , so her equation is $2 x + 2 y \frac{d y}{d x} + y + x \frac{d y}{d x} = 7$ . Explain her mistake, and find the correct gradient at $(1, 2)$ .
[4 marks]
Answer
C3
For the curve $x^{2} + 4 x y + 5 y^{2} = 4$ , Chen finds $\frac{d y}{d x} = - \frac{x + 2 y}{2 x + 5 y}$ . He says: "The stationary points are where $x = - 2 y$ , so there are infinitely many." Explain his mistake, and find the stationary points.
[5 marks]
Answer
C4
A curve has equation $e^{2 y} + x y = 1 + x$ . Dev differentiates $e^{2 y}$ and gets $2 e^{2 y}$ . Explain his mistake, and find an equation of the tangent to the curve at the origin.
[5 marks]
Answer

Exam-style questions

Edexcel exam style

D1
A curve $C$ has equation $3 x^{2} - 2 x y + y^{2} = 9$ . The point $P (2, 3)$ lies on $C$ .
(a)
Find $\frac{d y}{d x}$ in terms of $x$ and $y$ .
[3 marks]
Answer
(b)
Find an equation of the normal to $C$ at $P$ . Give it in the form $a x + b y + c = 0$ , where $a$ , $b$ and $c$ are integers.
[4 marks]
Answer
(c)
The normal at $P$ meets $C$ again at $Q$ . Find the coordinates of $Q$ .
[4 marks]
Answer
Total for D1: 11 marks

D2
A curve has equation $x^{2} + 4 y^{2} - 2 x y = 12$ . Ravi differentiates and writes: " $2 x + 8 y \frac{d y}{d x} - 2 y = 0$ , so $\frac{d y}{d x} = \frac{y - x}{4 y}$ ."
Ravi is wrong.
(a)
Explain Ravi's mistake, and find the correct $\frac{d y}{d x}$ .
[3 marks]
Answer
(b)
Find the coordinates of the stationary points on the curve.
[3 marks]
Answer
(c)
Find the coordinates of the points where the tangent to the curve is vertical.
[4 marks]
Answer
Total for D2: 10 marks
D3
A curve has equation $2^{x} + 2^{y} = 6$ . Find the gradient of the curve at the point $(2, 1)$ .
[6 marks]
Answer
D4
A curve has equation $x^{2} tan y = 9$ , where $x > 0$ and $0 < y < \frac{π}{2}$ .
(a)
Show that $\frac{d y}{d x} = - \frac{18 x}{x ^{4} + 81}$ .
[4 marks]
(b)
Find an equation of the tangent to the curve at the point where $x = 3$ . Give it in the form $y = m x + c$ , where $m$ and $c$ are exact.
[4 marks]
Answer
Total for D4: 8 marks
D5
The diagram shows the curve with equation $x^{3} + y^{3} = 9 x y$ . The point $P (2, 4)$ lies on the curve.
(a)
Find the gradient of the curve at $P$ .
[4 marks]
Answer
(b)
The normal to the curve at $P$ meets the $x$ -axis at $A$ and the $y$ -axis at $B$ . Find the area of triangle $O A B$ .
[5 marks]
Answer
Total for D5: 9 marks
D6
This question is about the curve $y = 4^{x} - 2^{x + 1}$ .
(a)
By taking logarithms of $y = 4^{x}$ , show that $\frac{d}{d x} (4^{x}) = 4^{x} ln 4$ .
[2 marks]
(b)
Find the coordinates of the stationary point on the curve $y = 4^{x} - 2^{x + 1}$ .
[3 marks]
Answer
(c)
Find the exact gradient of the curve at the point where $x = 1$ . Give it in the form $k ln 2$ .
[3 marks]
Answer
Total for D6: 8 marks

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
The diagram shows the curve with equation $x^{2} - x y + y^{2} = 3$ . Find the greatest value of $y$ on the curve. Then find the equations of the two tangents to the curve that are parallel to the line $y = x$ .
[9 marks]
Hint 1 · What to try
Differentiate implicitly: $\frac{d y}{d x} = \frac{2 x - y}{x - 2 y}$ .
Hint 2 · The first line
At the greatest $y$ the tangent is horizontal, so $y = 2 x$ . Put that back into the curve.
Hint 3 · The full method
Parallel to $y = x$ means $\frac{d y}{d x} = 1$ . That gives $x = - y$ ; put it back into the curve.
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can differentiate a term in y with respect to x by multiplying by dy/dx.
I can use the product rule on a term such as xy when differentiating implicitly.
I can find gradients, tangents, normals and stationary points of a curve given implicitly.
I can differentiate an exponential with any base, such as 3 to the power x.

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