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Differentiating : Differentiating : eexx, lnx and the chain , lnx and the chain
rulerule
© Christine Crisp
Objectives
To understand why the differentials of lnx and ex are what they are!
To be able to differentiate more complex functions that are composite functions
),( 42Tangent at
2xy (2, 4)x
The Gradient at a point on a CurveDefinition: The gradient of a point on a curve
equals the gradient of the tangent at that point.e.g.
3
12
The gradient of the tangent at (2, 4) is
43
12 mSo, the gradient of the curve at (2, 4)
is 4
2xy
2m
The Rule for Differentiation
1. The gradient of this curve at any point is the tangent
2. The tangent changes at any one point for a curve
2xy
4m
2xy
6m
The gradient Dee-Y by Dee-XThe gradient for a function is defined as
h
xfxfLimit
)()(0
Use this to show that the gradient of y=x2 is 2x
xxf
xfx
2|
2
2 minutes for this !
Solution Like this....
02
2
2
2
)(0
2
222
22
asx
x
x
xxx
xxLim
Differentiate ex and ln x
From first principles
This means use the definition of gradient as a limit
1 minute to think about this
Let’s consider the derivative of the exponential function. Going back to our limit definition of the derivative:
h
eelime
dx
d xhx
0h
x
First rewrite the exponential using exponent rules.
h
eeelim
xhx
0h
Next, factor out ex. h
1eelim
hx
0h
Since ex does not contain h, wecan move it outside the limit.
h
xfhxflim
dx
d0h
xf
h
1elime
h
0h
x
Substituting h=0 in the limit expression results in the
indeterminate form , thus we will need to determine it.
00
We can look at the graph of and observe what happens as x gets close to 0. We can also create a table of values close to either side of 0 and see what number we are closing in on.
x1e
xfx
)(
x -.1 -.01 -.001 .001 .01 .1 y .95 .995 .999 1.0005 1.005 1.05
Graph
At x = 0, f(0) appears to be 1.
Table
As x approaches 0, y approaches 1.
X≠0 of course
We can safely say that from the last slide that 1
h
1elim
h
0h
Thus
Rule 1: Derivative of the Exponential Function
xxh
0h
xx e1eh
1elimee
dx
d
xx eedx
d
The derivative of the exponential function is the exponential function.
What about ln x
Method 1 – Using previous result
Method 2 – Consider Limits
What about ln x
Method 1 – Using previous result
Method 2 – Consider Limits
y = ln x
therefore ey = x ( raise both sides to e) dx/dy = ey. (Oh yes it does!!)
Since dx/dy * dy/dx = 1 1/(dx/dy) = dy/dx.
= 1/ey f'(x) = 1/x. (as ey=x)
y = ln x
f(x) = ln x f(x+h) = ln (x+h) f'(x) = lim(h->0) [ln (x+h) - ln x] / h
= lim(h->0) ln (x+h/x) / h
= lim(h->0) 1/h * ln(1+h/x) Since lim(h->0) ln(1+h/x) -> h/x where h ≠ 0,
f'(x) = 1/h * h/x = 1/x
Exercise
The Chain Rule
The gradient at a point on a curve is defined as the gradient of the tangent at that point.
The process of finding the gradient function is called differentiating.
The function that gives the gradient of a curve at any point is called the gradient function.
The rules we have developed for differentiating are:
A reminder of the differentiation done so far!
1 nn nxdx
dyxy
axax aedx
dyey
The Chain Rule
dx
dyxxy 168 36
We can find by multiplying out the brackets: dx
dy
However, the chain rule will get us to the answer without needing to do this ( essential if we had, for example, . )103 )4( x
)4(6 32 xx
2)( xxf Suppose and 4)( 3 xxg
))(( xgfyLet23 )4( x
25 246 xx
Differentiating a function of a function .
))(( xgf
)4( 3xf
The Chain Rule
2)( xxf Consider again and .4)( 3 xxg
Let 43 xuDifferentiating both these expressions:
2uy Then,
We must get the letters right.
23xd
d
ux
ud
d2
y
u
))(( xgfyLet 23 )4( x
The Chain Rule
2)( xxf Consider again and .4)( 3 xxg
Let 43 xuDifferentiating both these expressions:
2uy Then,
23xd
d
ux
ud
d2
y
u)4(2 3 x
))(( xgfyLet 23 )4( x
Now we can substitute for u
The Chain Rule
2)( xxf Consider again and .4)( 3 xxg
Let 43 xuDifferentiating both these expressions:
2uy Then,
23xd
d
ux
ud
d2
y
u)4(2 3 x
))(( xgfyLet 23 )4( x
Can you see how to get to the answer which we know is
?)4(6 32 xxdx
dy
We need to multiply by23x )4(2 3 x
The Chain Rule
23xdx
du u
du
dy2 )4(2 3 x
So, we have
So,
dx
du
du
dy
dx
dy
and to get we need to multiply by23x )4(2 3 xdx
dy
This expression is behaving like fractions with the ‘s on the r,h,s, cancelling.
du
The Chain Rule
23xdx
du u
du
dy2 )4(2 3 x
So, we have
So,
dx
du
du
dy
dx
dy
and to get we need to multiply by23x )4(2 3 xdx
dy
This expression is behaving like fractions with the s on the r,h,s, cancelling.
du
Although these are not fractions, they come from taking the limit of the gradient, which is a fraction.
The Chain Rule
Solution:
We need to recognise the function as and identify the inner function ( which is u ).
))(( xgf)(xg
5)(ye.g. 1 Find ifdx
dyx41
The Chain Rule5)(y
4dx
du
e.g. 1 Find if
dx
du
du
dy
dx
dy
dx
dy
45udu
dy
Solution:
x41
xu 41 Let Then 5uy
4)41(5 4 xdx
dy
4)41(20 x
Differentiating:
We need to recognise the function as and identify the inner function ( which is u ).
))(( xgf)(xg
We don’t multiply out the brackets
4)41(5 xSubstitute for u
Tidy up by writing the constant first
The Chain Rule
xy
52
1
e.g. 2 Find if
dx
dy
Solution: We can start in 2 ways. Can you spot them?
21
)52(
xyEither write
xu 52 and then let
Or, if you don’t notice this, start with xu 52
Thenu
y1
21
1
u
21
u
so 2
1uy
Always use fractions for indices, not decimals.
The Chain Rule
xy
52
1
e.g. 2 Find if
dx
dy
5dx
du
dx
du
du
dy
dx
dy
23
2
1 u
du
dy
xu 52
5)52(2
123
xdx
dy
23
)52(2
1 x
Solution: Whichever way we start we get
23
)52(2
5 x
21
uyand
The Chain Rule
The chain rule is used for differentiating functions of a function.
dx
du
du
dy
dx
dy
)(xgu where , the inner function.
If ,))(( xgfy
SUMMARY
The Chain RuleExercis
e
dx
dyUse the chain rule to find for the following:
32 )3( xy
1.
2.
8)21( xy
3.
43 xy4.
5.
51
1
xy
xy
31
2
Solutions:
1.
32 3 uyxu
xdx
du2
22 )3(6 xxdx
dy
222 )3(33 xudu
dy
dx
du
du
dy
dx
dy
The Chain RuleSolutio
ns2.
8)21( xy
3.
21
)43(43 xxy
821 uyxu
2dx
du 77 )21(88 xudu
dy
7)21(16 xdx
dy
21
43 uyxu
3dx
du21
21
)43(2
1
2
1 xu
du
dy
21
)43(2
3 x
dx
dy
dx
du
du
dy
dx
dy
dx
du
du
dy
dx
dy
The Chain RuleSolutions
4.
1)31(231
2
xx
y
1231 uyxu
3dx
du 22 )31(22 xudu
dy
2)31(6 xdx
dydx
du
du
dy
dx
dy
The Chain RuleSolutions5.
51
1
xy
4
2
11
5
xxdx
dy
511 uy
xu
22 1
xx
dx
du
44 1
155
xu
du
dy
dx
du
du
dy
dx
dy
11 x
The Chain Rule
The chain rule can also be used to differentiate functions involving e.
xedx
dy 22
e.g. 3. Differentiate xey 2
Solution: The inner function is the 1st operation on x, so u = 2x.
Let xu 2 uey
2dx
du xu eedu
dy 2
dx
du
du
dy
dx
dy
The Chain RuleExercis
eUse the chain rule to differentiate the
following: xey 3
1.
2.
xey
1
3.
2xey
Solutions:
1.
ueyxu 3
3dx
du
xedx
dy 33
xu eedu
dy 3
dx
du
du
dy
dx
dy
The Chain RuleSolutio
ns2.
xx
ee
y 1
3.
2xey
xu eedu
dy
ueyxu
1dx
du
ueyxu 2
xdx
du2
2xu eedu
dy
22 xxe
dx
dy
dx
du
du
dy
dx
dy
dx
du
du
dy
dx
dy
xedx
dy
The Chain Rule
Later we will want to reverse the chain rule to integrate some functions of a function.To prepare for this, we need to be able to use the chain rule without writing out all the steps.
32 )3( xye.g. For we know that 22 )3(6 xxdx
dy
has been multiplied by the derivative of
the outer function3)_____(
The derivative of the inner function which is
)3( 2 xx2
( I’ve put dashes here because we want to ignore the inner function at this stage. We must not differentiate it again. )
2)_____(3which is
The Chain Rule
So, the chain rule says
differentiate the inner function
multiply by the derivative of the outer function
e.g. dx
dyy
x4e
The Chain Rule
So, the chain rule says
differentiate the inner function
multiply by the derivative of the outer function
4e.g. dx
dyy
x4
( The inner function is ) x4
e
The Chain Rule
So, the chain rule says
differentiate the inner function
multiply by the derivative of the outer function
4 xe 4
( The outer function is )e
( The inner function is )
e.g. dx
dyy
x4e
x4
The Chain Rule
So, the chain rule says
differentiate the inner function
multiply by the derivative of the outer function
4
xe 44
xe 4
( The outer function is )
( The inner function is )
e.g. dx
dyy
x4e
x4e
The Chain RuleBelow are the exercises you have already done using the chain rule with exponential functions.
xey 3
1.
2.
xey
1
3.
2xey
See if you can get the answers directly.
Answers:
xedx
dy 33
1.
2.
xedx
dy
3.
22 xxe
dx
dy
Notice how the indices never change.
xe
The Chain Rule
TIP: When you are practising the chain rule, try to write down the answer before writing out the rule in full. With some functions you will find you can do this easily.However, be very careful. With some functions it’s easy to make a mistake, so in an exam don’t take chances. It’s probably worth writing out the rule.
The Chain Rule
The chain rule is used for differentiating functions of a function.
dx
du
du
dy
dx
dy
)(xgu where , the inner function.
If ,))(( xgfy
SUMMARY
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