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Derivatives of Logarithmic
Functions

John Napier (1550–1617), the
inventor of logarithms
4.2 derivatives of logarithmic functions
d
= 1 ,
log bx  x ln b
dx

x>0

d
1
x>0
[ ln x ] = ,
dx
x
d
1 du
x>0
[ ln u] = × ,
dx
u dx

Pg.244 deriving
Pg.244 deriving
formulas from
formulas from
definition
definition
Example: Find the derivative.

(

)

f ( x ) = ln x 2 + 3

1
f ' ( x) = 2
×2x
( x + 3)

2x
f '( x) = 2
( x + 3)

Example: Find the derivative.

x
f ( x ) = ln
x +1



1 1× x +1) −1×x ÷
(
f '( x) =
×
÷
x  ( x +1) 2


x +1
x +1  ( x +1) − x 

÷
f '( x) =
×
2 ÷
x  ( x +1) 


f ' ( x) =

x +1− x
x ( x +1)

1
f '( x) =
x ( x +1)
One more example !!!
  x 2 sin x 
f (x) = ln 
÷
  1+ x 

Find the derivative of



1

 2x sin x + (−cos x)×x 2  × 1+ x −
×x 2 sin x


1
2 1+ x
×
f '(x) =  2
2
  x sin x  
1+ x
÷

  1+ x  

(

)

x 2 sin x
2x sin x − x cos x) × 1+ x −


2 1+ x
f '(x) =
x 2 sin x 1+ x
2

(

)
Lets try this again
Find the derivative of

  x 2 sin x 
f (x) = ln 
÷
  1+ x 

1
  x 2 sin x 
2
2
f (x) = ln 
÷ = ln ( x sin x ) − ln 1+ x = ln x + lnsin x − ln ( 1+ x ) 2
  1+ x 

therefore

1
f (x) = 2 ln x + ln ( sin x ) − ln ( 1+ x )
2
so…

1
1
1 1
f '(x) = 2 × +
×cos x − ×
x sin x
2 1+ x
2
1
f '(x) = + cot x −
x
2 ( 1+ x )
Logarithmic Differentiation
The derivative of

y=

x 2 3 7x −14

( 1+ x )
2

4

is to messy to calculate directly

so we first take the natural logarithm of both sides and use its properties.

(

ln y = ln x

23

)

7x −14 − ln ( 1+ x

)

2 4

1
3

= ln x + ln ( 7x −14) − ln ( 1+ x

therefore

2

1
2
ln y = 2 ln x + ln ( 7x −14) − 4 ln ( 1+ x )
3

)

2 4
1 dy
1 1
1
1
=2 +
×7 − 4
×2x
2
y dx
x 3 ( 7x −14)
( 1+ x )
1 dy 2
7
8x
= +
−
y dx x 3 ( 7x −14) ( 1+ x 2 )


1 dy  2
7
8x ÷
y×
= +
−
×y
2 ÷
y dx  x 3 ( 7x −14) ( 1+ x ) 


 23
dy  2
7
8x ÷ x 7x −14
= +
−
×
2 ÷
 x 3 ( 7x −14) ( 1+ x )
2 4
dx 
 ( 1+ x )

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4.2 derivatives of logarithmic functions

  • 1. Derivatives of Logarithmic Functions John Napier (1550–1617), the inventor of logarithms
  • 3. d = 1 , log bx  x ln b dx x>0 d 1 x>0 [ ln x ] = , dx x d 1 du x>0 [ ln u] = × , dx u dx Pg.244 deriving Pg.244 deriving formulas from formulas from definition definition
  • 4. Example: Find the derivative. ( ) f ( x ) = ln x 2 + 3 1 f ' ( x) = 2 ×2x ( x + 3) 2x f '( x) = 2 ( x + 3) Example: Find the derivative. x f ( x ) = ln x +1   1 1× x +1) −1×x ÷ ( f '( x) = × ÷ x  ( x +1) 2   x +1 x +1  ( x +1) − x   ÷ f '( x) = × 2 ÷ x  ( x +1)   f ' ( x) = x +1− x x ( x +1) 1 f '( x) = x ( x +1)
  • 5. One more example !!!   x 2 sin x  f (x) = ln  ÷   1+ x  Find the derivative of   1   2x sin x + (−cos x)×x 2  × 1+ x − ×x 2 sin x   1 2 1+ x × f '(x) =  2 2   x sin x   1+ x ÷    1+ x   ( ) x 2 sin x 2x sin x − x cos x) × 1+ x −   2 1+ x f '(x) = x 2 sin x 1+ x 2 ( )
  • 6. Lets try this again Find the derivative of   x 2 sin x  f (x) = ln  ÷   1+ x  1   x 2 sin x  2 2 f (x) = ln  ÷ = ln ( x sin x ) − ln 1+ x = ln x + lnsin x − ln ( 1+ x ) 2   1+ x  therefore 1 f (x) = 2 ln x + ln ( sin x ) − ln ( 1+ x ) 2 so… 1 1 1 1 f '(x) = 2 × + ×cos x − × x sin x 2 1+ x 2 1 f '(x) = + cot x − x 2 ( 1+ x )
  • 7. Logarithmic Differentiation The derivative of y= x 2 3 7x −14 ( 1+ x ) 2 4 is to messy to calculate directly so we first take the natural logarithm of both sides and use its properties. ( ln y = ln x 23 ) 7x −14 − ln ( 1+ x ) 2 4 1 3 = ln x + ln ( 7x −14) − ln ( 1+ x therefore 2 1 2 ln y = 2 ln x + ln ( 7x −14) − 4 ln ( 1+ x ) 3 ) 2 4
  • 8. 1 dy 1 1 1 1 =2 + ×7 − 4 ×2x 2 y dx x 3 ( 7x −14) ( 1+ x ) 1 dy 2 7 8x = + − y dx x 3 ( 7x −14) ( 1+ x 2 )   1 dy  2 7 8x ÷ y× = + − ×y 2 ÷ y dx  x 3 ( 7x −14) ( 1+ x )     23 dy  2 7 8x ÷ x 7x −14 = + − × 2 ÷  x 3 ( 7x −14) ( 1+ x ) 2 4 dx   ( 1+ x )