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““Teach A Level Maths”Teach A Level Maths”

Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules

xy ln6: 6: Differentiating Differentiating

xy ln

Differentiating xy ln

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Module C3

Differentiating xy lnmeans

xy ln xy elog

yex

A log is just an index, so

xy elog

So, to differentiate we just need to differentiate .

xy lnyex

We have to be careful with the letters: x and y have swapped from their usual

places.

xSo,y e

Differentiating xy lnmeans

xy ln xy elog

yex

A log is just an index, so

xy elog

So, to differentiate we just need to differentiate .

xy lnyex

We have to be careful with the letters: x and y have swapped from their usual

places.

xSo,y e

d

d

y

x ye

Differentiating xy lnmeans

xy ln xy elog

yex

A log is just an index, so

xy elog

So, to differentiate we just need to differentiate .

xy lnyex

We have to be careful with the letters: x and y have swapped from their usual

places.

However, for , we want notxy lndx

dydy

dx

xSo,y e

d

d

y

x ye

Differentiating xy ln

We’ve already seen that behaves like a fraction, dx

dy

dx

dy

dy

dx

1

yedx

dy 1 ye

dy

dxHence,

xy lnxdx

dy 1So, since we getyex

Finally, for , we want the answer in terms of x. dx

dy

So,Compare this with

2

3

1

3

2

xy lnxdx

dy 1

SUMMARY

Differentiating )(ln xfy Compound Functions Involving logsWe can always use the chain rule to differentiate compound log functions.

However, the first 3 log laws met in AS can simplify the work.

Using ln instead of log these are:

xyln yx lnln

y

xln

kxln

yx lnln

xk ln

It’s important to use these laws as they change compound functions into simple ones.

Differentiating )(ln xfy

x

1

e.g.1 Use log laws to simplify the following

and hence find . dx

dy

Solution:

(a)

xy 3ln 2ln xy 2

lnx

y (b)

(c)

(a)

xy 3ln

x

10

(b)

2ln xy

xy ln3ln

xy ln2x

12

x

2

(c)

2

lnx

y 2lnln xyxdx

dy 1

dx

dy

is a constant so its derivative is zero

3lndx

dy

Differentiating )(ln xfy

It may seem surprising that the gradient

functions of and xy 3ln,xy ln2

lnx

y

are all given by xdx

dy 1

The graphs show us why:

xxy ln3ln3ln is a translation from

,xy ln

of

3ln

0 . So, we have

Differentiating )(ln xfy

xxy ln3ln3ln

2ln

0,xy ln

Similarly, is a translation from

of

2lnln2

ln xx

y

xy 3lnxy ln

Differentiating )(ln xfy

2lnln2

ln xx

y

xy 3ln

2lnx

y

xy ln

Since the graphs are translations parallel to the y-axis, the gradients are the same.

xxy ln3ln3ln

Differentiating )(ln xfy

Solution:

xu 21

e.g.2 Differentiate with respect to

x.)21ln( xy

(a) cannot be simplified. There’s no rule for the log of a sum or difference.

)21ln( xy

We must use the chain rule.uy ln

2dx

du

udu

dy 1

x21

1

xdx

dy

21

2

xdx

dy

21

12

So,

Differentiating )(ln xfy

We can generalise this to get a really useful result

dx

dyxfy )(ln

)21ln( xy xdx

dy

21

2

So,

)(

)(

xf

xf

where is the derivative of the function)(xf )(xf

“ The derivative of the inner function divided by the inner function.”

This rule in words is:

Differentiating )(ln xfy

Solution:

e.g.3 Differentiate

x

xy

1

1ln

xdx

dy

1

1

)1ln()1ln( xxy

x

xy

1

1ln

We can now differentiate each term separately, using the result from the last example:

dx

dyxfy )(ln

)(

)(

xf

xf

)1ln()1ln( xxy

So,

xxdx

dy

1

1

1

1x

1

1

The brackets are essential here.

Differentiating )(ln xfy SUMMARYTo differentiate compound log

functions, Use the log laws to simplify the

expression if possible. Differentiate each term

using

dx

dyxfy )(ln

)(

)(

xf

xf

“ The derivative of the inner function divided by the inner function.”

This rule in words is:

Differentiating )(ln xfy Exercises

Differentiate the following with respect to x:

1. xy 2ln 2.3

1ln

xy 3. 3ln xy

4.x

xy

21

21ln

5.)1)(1ln( xxy

1. xy 2ln

Solutions:

xy ln2ln xdx

dy 1

2.3

1ln

xy 3ln)1ln( xy

1

1

xdx

dy

Differentiating )(ln xfy Exercises

x

xy

21

21ln

5.

4. )1)(1ln( xxy

xdx

dy 33. 3ln xy xy ln3

)1ln()1ln( xxy

xxdx

dy

1

1

1

1

xxdx

dy

1

1

1

1

)21ln()21ln( xxy

xxdx

dy

21

2

21

2

xxdx

dy

21

2

21

2

Differentiating xy ln

Differentiating xy ln

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

Differentiating xy ln

xy lnxdx

dy 1

To differentiate compound log functions,

Use the log laws to simplify the expression if possible.

Differentiate each term using

dx

dyxfy )(ln

)(

)(

xf

xf

“ The derivative of the inner function divided by the inner function.”

This rule in words is:

Differentiating xy ln

x

1

e.g.1 Use log laws to simplify the following

and hence find . dx

dy

Solution:

(a)

xy 3ln 2ln xy 2

lnx

y (b)

(c)

(a)

xy 3ln

x

10

(b)

2ln xy

xy ln3ln

xy ln2x

12

x

2

(c)

2

lnx

y 2lnln xyxdx

dy 1

dx

dy

dx

dy

Differentiating xy ln

Solution:

e.g. Differentiatex

xy

1

1ln

xdx

dy

1

1

)1ln()1ln( xxy

x

xy

1

1ln

We can now differentiate each term separately, using:

dx

dyxfy )(ln

)(

)(

xf

xf

)1ln()1ln( xxy

So,

xxdx

dy

1

1

1

1x

1

1