DIFFEENTIATION OF LOGARITHMIC FUNCTIONS
Let a is a real number (>1)
If a^{x} = b
Then this can written as
Log_{a}b =X
And read as log of b to the base a is equal to X
So we can say that
, a^{x} =b Û log_{a}b =X
In logarithm base is very important
Logarithm of same number to the different bases are different
2^{6 } =64 Ûlog_{2} 64 =6
4^{3} =64Ûlog_{4}64 =3
8^{2} =64Ûlog_{8}64 =2
64^{1} =64Ûlog_{64}64 =1
Here we find that logarithm of same number (64) is different when bases are different
On this basis we having two types of logarithms
 Common logarithm with base 10 or other bases
 Natural logarithms with base e
Here are some formulas helpful in derivatives of logarithmic functions (at any base)
1.log xy = logx + logy
2. logx/y = logx – logy
3.logx^{n} =nlog x
4.log_{a}b × log_{b}a =1
5.log_{a}a =1
6.log1= 0
Derivative of log
 d(logx)/dx =1/x

d(log_{a}x)/dx =1/xloga
where we use log to differentiate
 When product or division of many functions are given

When power is a function or a variable
In this conditions we take log both sides,then differentiate
Some examples
1. If Y =x^{x} find dy/dx
Solution
Y = x^{x}
Taking log both sides
Log y = log x^{x}
Logy = xlogx
Differentiating both sides w.r.t.x ,we get
1/y( dy/dx) =x(dlogx/dx) + logx dx/dx
dy/dx =y (x/x +logx)
dy/dx = x^{x} (1+logx)
2.If y=log log logx^{3},find dy/dx
Solution
Y =log log logx^{3}
dy/dx =d(log log log x^{3})/d(log logx^{3}) x d(loglogx^{3}) /d(logx^{3}) x d(logx^{3}) /dx^{3}/dx
dy/dx = (1/log log x^{3} ) x (1/ logx^{3}) x(1/x^{3}) x 3x^{2}
dy/dx = 3x^{2}/ (log logx^{3}) x (logx^{3}) x (x^{3})^{ }

IF y=tanx tan2x tan3x tan4x find dy/dx
Y =tanx tan2x tan3x tan4x
Taking log both sides
logy =logtanx +logtan2x +logtan3x +logtan4x
differentiating both sides w.r.t.x,
1/y dy/dx =sec^{2}x/tanx +2sec^{2}2x/tan2x +3sec^{2}3x/tan3x +4sec^{2}4x/tan4x
=) x (cosec2x +2cosec4x +3cosec6x +4cosec8x)