differentiation 12 implicit differentation...

Differentiation 12implicit differentation explanations

J A Rossiter

1

Slides by Anthony Rossiter

Introduction

• So far these resources have focussed on cases where the function is simply defined as y=f(x), w=g(z) or similar.

• That is a single explicit output and a single independent variable.

• Sometimes the relationship between 2 variables is not explicit and cannot be solved algebraically, for example:

How do we find the derivative of such a curve?


2

)1cos()sin( 2 xxyy

Observation

dy/dx is the gradient of the curve with x on the horizontal axis and y on the vertical axis.

A simple swapping of the axis reveals that the gradient is of course the inverse.


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3 3.5 4 4.5 5 5.5 6-400

-200

0

200

400

600

800

1000

1200

x

Gradient is dy/dx=470

Gradient is dx/dy=1/470

)(xfy

)(1 yfx

Assumption

Consider a function written as:

Assume the inverse function exists, so that we can easily differentiate to find dx/dy:

It is obvious that the gradients of each of these curves are the inverses one of another, as in essence they are the same curve but with swapped axis.


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)(xfy

1.1

dy

dx

dx

dyor

dy

dxdx

dy

In fact, students can guess that this must be

true, even if a simple inverse function does

not exist.

)(1 yfx

Simple logarithm function

Find the derivative of the function

STEP 1: Write x as a function of y (the inverse function).

STEP 2: Differentiate wrt to y.

STEP 3: Use inverse to find dy/dx.


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)log(xy

yex ye

dy

dx

xedy

dxdx

dyy

111

Inverse sine

Find the gradient dy/dx of a curve defined with the following expression.

METHOD: First use the inverse function.

Now differentiate wrt x.


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axy 1sin

dy

dx

a

y

)cos(

22 )(1sin1)cos( ax

a

y

a

y

a

dx

dy

xa

y

)sin(

Using the chain rule on y

Let the expression contain functions of the ‘dependent variable’ y.

Using the chain rule the following is obvious.

Easy examples:


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dx

dy

dy

dfyf

dx

d.))((

dx

dyyy

dx

d.2)( 2

dx

dyyy

dx

d).cos())(sin(

Implicit differentiation

We will make use of two observations:

• The fact that (dy/dx) . (dx/dy) =1

• The chain rule.

Using these we can differentiate an expression containing two variables and separate out a derivative.


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Example 1

Find the gradient of a curve defined with the following expression.

METHOD: Differentiate every term in the expression, using the chain rule for any terms which include a ‘y’.

Separate variables:


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xxyy 3)2cos(

13)2sin(2 2 xdx

dyy

dx

dy

)2sin(21

13 2

y

x

dx

dy

NOTE: The answer includes values from both

x and y!

Example 2 – a circle

Find the gradient of a curve defined with the following expression.


Separate variables:


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422 xy

022 xdx

dyy

y

x

dx

dy

NOTE: The answer includes values from

both x and y!

Circle

It is easy to show that this formulae works.


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422 xy

y

x

dx

dy

3

1

3

1

dx

dy

y

x

Example 2 – observation

In this case one could write down an expression for y as follows.

Hence one can deduce:

Same answer as on the previous page:


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44 222 xyxy

y

x

dx

dy

42

x

x

dx

dy

Example 3 – back to front



Separate variables:


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xyyy 4)cos(42

4)]sin(42[ 3 dx

dyyyy

)]sin(42[

43 yyydx

dy

Example 3 – alternative


METHOD: Differentiate with respect to y and then note the relationship between dy/dx and dx/dy.


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xyyy 4)cos(42

dy

dxyyy 4)]sin(42[ 3

)]sin(42[

413 yyy

dydxdx

dy

Example 4 – logarithms


METHOD: First express using exponentials.

Now differentiate every term.


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)13log( 2 xxy

32 xdx

dye y

13

32322

xx

x

e

x

dx

dyy

132 xxe y

Observation


METHOD: First express using exponentials.

Now differentiate every term.


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))(log( xfy

dx

df

dx

dye y

)(xf

dxdf

e

dxdf

dx

dyy

)(xfe y

Summary• This brief resource has derived the rule for implicit

differentiation.

• This is needed when the output or dependent variable is not defined explicitly in terms of the independent variable and is also useful for inverse functions.

• The basic method is to differentiate every term in the expression, using the chain rule as required, and then separate out the terms which include the derivative.

• At times recognising (dy/dx ). (dx/dy) =1 can be useful.


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© 2016 University of Sheffield

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Anthony RossiterDepartment of Automatic Control and

Systems EngineeringUniversity of Sheffieldwww.shef.ac.uk/acse

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