twelve

Implicit
Differentiation
mc-TY-implicit-2009-1
Sometimes functions are given not in the form y = f(x) but in a more complicated form in which
it is difficult or impossible to express y explicitly in terms of x. Such functions are called implicit
functions. In this unit we explain how these can be differentiated using implicit differentiation.
In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature.
After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
• differentiate functions defined implicitly
Contents
1. Introduction 2
2. Revision of the chain rule 2
3. Implicit differentiation 4
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1. Introduction
In this unit we look at how we might differentiate functions of y with respect to x.
Consider an expression such as
x2 + y2
? 4x + 5y ? 8 = 0
It would be quite difficult to re-arrange this so y was given explicitly as a function of x. We could
perhaps, given values of x, use the expression to work out the values of y and thereby draw a
graph. In general even if this is possible, it will be difficult.
A function given in this way is said to be defined implicitly. In this unit we study how to
differentiate a function given in this form.
It will be necessary to use a rule known as the the chain rule or the rule for differentiating a
function of a function. In this unit we will refer to it as the chain rule. There is a separate unit
which covers this particular rule thoroughly, although we will revise it briefly here.
2. Revision of the chain rule
We revise the chain rule by means of an example.
Example
Suppose we wish to differentiate y = (5 + 2x)10 in order to calculate dy
dx .
We make a substitution and let u = 5 + 2x so that y = u10.
The chain rule states
dy
dx
=
dy
du
×
du
dx
Now
if y = u10 then
dy
du
= 10u9
and
if u = 5 + 2x then
du
dx
= 2
hence
dy
dx
=
dy
du
×
du
dx
= 10u9
× 2
= 20u9
= 20(5 + 2x)9
So we have used the chain rule in order to differentiate the function y = (5 + 2x)10.
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In quoting the chain rule in the form
dy
dx
=
dy
du
×
du
dx
note that we write y in terms of u, and u
in terms of x. i.e.
y = y(u) and u = u(x)
We will need to work with different variables. Suppose we have z in terms of y, and y in terms
of x, i.e.
z = z(y) and y = y(x)
The chain rule would then state:
dz
dx
=
dz
dy
×
dy
dx
Example
Suppose z = y2. It follows that dz
dy = 2y. Then using the chain rule
dz
dx
=
dz
dy
×
dy
dx
= 2y ×
dy
dx
= 2y
dy
dx
Notice what we have just done. In order to differentiate y2 with respect to x we have differentiated
y2 with respect to y, and then multiplied by
dy
dx
, i.e.
d
dx ??y2
=
d
dy ??y2
×
dy
dx
We can generalise this as follows:
to differentiate a function of y with respect to x, we differentiate with respect to y and then
multiply by dy
dx .
Key Point
d
dx
(f(y)) =
d
dy
(f(y)) ×
dy
dx
We are now ready to do some implicit differentiation. Remember, every time we want to differentiate
a function of y with respect to x, we differentiate with respect to y and then multiply by
dy
dx .
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3. Implicit differentiation
Example
Suppose we want to differentiate the implicit function
y2 + x3
? y3 + 6 = 3y
with respect x.
We differentiate each term with respect to x:
d
dx ??y2
+
d
dx ??x3
?
d
dx ??y3
+
d
dx
(6) =
d
dx
(3y)
Differentiating functions of x with respect to x is straightforward. But when differentiating a
function of y with respect to x we must remember the rule given in the previous keypoint. We
find
d
dy ??y2
×
dy
dx
+ 3x2
?
d
dy ??y3
×
dy
dx
+ 0 =
d
dy
(3y) ×
dy
dx
that is
2y
dy
dx
+ 3x2
? 3y2 dy
dx
= 3
dy
dx
We rearrange this to collect all terms involving dy
dx together.
3x2 = 3
dy
dx
? 2y
dy
dx
+ 3y2 dy
dx
then
3x2 = ??3 ? 2y + 3y2

dy
dx
so that, finally,
dy
dx
=
3x2
3 ? 2y + 3y2
This is our expression for
dy
dx
.
Example
Suppose we want to differentiate, with respect to x, the implicit function
sin y + x2y3
? cos x = 2y
As before, we differentiate each term with respect to x.
d
dx
(sin y) +
d
dx ??x2y3
?
d
dx
(cos x) =
d
dx
(2y)
Recognise that the second term is a product and we will need the product rule. We will also use
the chain rule to differentiate the functions of y. We find
d
dy
(sin y) ×
dy
dx
+ x2 d
dx ??y3
+ y3 d
dx ??x2
 + sin x =
d
dy
(2y) ×
dy
dx
so that
cos y
dy
dx
+ x2.
d
dy ??y3

dy
dx
+ y3. 2x + sin x = 2
dy
dx
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Tidying this up gives
cos y
dy
dx
+ x2 3y2 dy
dx
+ 2xy3 + sin x = 2
dy
dx
We now start to collect together terms involving dy
dx .
2xy3 + sin x = 2
dy
dx
? cos y
dy
dx
? 3x2y2 dy
dx
2xy3 + sin x = (2 ? cos y ? 3x2y2)
dy
dx
so that, finally
dy
dx
=
2xy3 + sin x
2 ? cos y ? 3x2y2
We have deliberately included plenty of detail in this calculation. With practice you will be able
to omit many of the intermediate stages. The following two examples show how you should aim
to condense the solution.
Example
Suppose we want to differentiate y2 + x3
? xy + cos y = 0 to find dy
dx . The condensed solution
may take the form:
2y
dy
dx
+ 3x2
?
d
dx
(xy) ? sin y
dy
dx
= 0
(2y ? sin y)
dy
dx
+ 3x2
? x
dy
dx
+ y.1 = 0
(2y ? sin y ? x)
dy
dx
+ 3x2
? y = 0
(2y ? sin y ? x)
dy
dx
= y ? 3x2
so that
dy
dx
=
y ? 3x2
2y ? sin y ? x
Example
Suppose we want to differentiate
y3
? x sin y +
y2
x
= 8
The solution is as follows:
3y2 dy
dx
? x cos y
dy
dx
+ sin y.1 +
x 2y dy
dx
? y2.1
x2 = 0
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Multiplying through by x2 gives:
3x2y2 dy
dx
? x3 cos y
dy
dx
? x2 sin y + 2xy
dy
dx
? y2 = 0
dy
dx ??3x2y2
? x3 cos y + 2xy
= x2 sin y + y2
so that
dy
dx
=
x2 sin y + y2
3x2y2
? x3 cos y + 2xy
Exercises
1. Find the derivative, with respect to x, of each of the following functions (in each case y
depends on x).
a) y b) y2 c) sin y d) e2y e) x + y
f) xy g) y sin x h) y sin y i) cos(y2 + 1) j) cos(y2 + x)
2. Differentiate each of the following with respect to x and find dy
dx .
a) sin y + x2 + 4y = cos x.
b) 3xy2 + cos y2 = 2x3 + 5.
c) 5x2
? x3 sin y + 5xy = 10.
d) x ? cos x2 + y2
x + 3x5 = 4x3.
e) tan 5y ? y sin x + 3xy2 = 9.
Answers to Exercises on Implicit Differentiation
1.
a)
dy
dx
b) 2y
dy
dx
c) cos y
dy
dx
d) 2e2y dy
dx
e) 1 +
dy
dx
f) x
dy
dx
+ y
g) y cos x + sin x
dy
dx
h) (sin y + y cos y)
dy
dx
i) ?2y sin(y2 + 1)
dy
dx
j) ?2y
dy
dx
+ 1sin(y2 + x)
2.
a)
dy
dx
=
?sin x ? 2x
4 + cos y
b)
dy
dx
=
6x2
? 3y2
6xy ? 2y sin y2
c)
dy
dx
=
10x ? 3x2 sin y + 5y
x3 cos y ? 5x
d)
dy
dx
=
12x4
? 15x6 + y2
? 2x3 sin x2
? x2
2xy
e)
dy
dx
=
y cos x ? 3y2
5sec25y ? sin x + 6xy
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