CK-12-Calculus

6.3. Differentiation and Integration of Logarithmic and Exponential Functions http://www.ck12.org

and in the special case whereb=e,

d

dx[lnx] =

1

x>^0.

To generalize, ifuis a differentiable function ofxand ifu(x)> 0 ,then the above two equations, after the Chain Rule
is applied, will produce the generalized derivative rule for logarithmic functions.
Derivatives of Logarithmic Functions

d

dx[logbu] =

1

ulnb

du

dx

d

dx[lnu] =

1

u

du

dx=

u′

u

Remark:Students often wonder why the constanteis defined the way it is. The answer is in the derivative of
f(x) =lnx.With any other base the derivative off(x) =logbxwould be equalf′(x) =xln^1 b,a more complicated
expression than 1/x.Thinking back to another unexpected unit, radians, the derivative off(x) =sin(x)is the simple
expression f′(x) =cos(x)only ifxis in radians. In degrees,f′(x) = 180 π cos(x), which is more cumbersome and
harder to remember.
Example 1:
Find the derivative ofy=ln( 2 x^2 − 4 x+ 3 ).
Solution:
Sincedxd[lnu] =^1 ududx, foru= 2 x^2 − 4 x+ 3 ,

dy

dx=

1

2 x^2 − 4 x+ 3

d

dx

[ 2 x (^2) − 4 x+ 3 ]
= 2 x (^2) −^14 x+ 3 ( 4 x− 4 )
= 2 x^42 (−x− 4 x^1 +) 3.
Example 2:
Finddxd[ln(sinx)].
Solution:
d
dx[ln(sinx)] =

1

sinx·[cosx]

=cossinxx

=cotx.

Example 3:
Finddxd[ln(cos 5x)^3 ].
Solution: