# derivative of 1)logx by first principle

let   f(x) = log x

f(x + h) = log (x +h)  {small increment }

by first principle

dy/dx  =   f(x + h) - f(x) / h  as  h tans to 0

=   log (x +h)  -  log x  / h  h tans to 0

=  log (x + h) /x whole divide by h  h tans to 0  {using log m -  log n  = log (m)/n }

=   log (1+h/x)  /  h       h tans to 0

=   log (1+h/x)  /  xh/x   h tans to 0  {log (1+h/x)  / h/x   h tans to 0  = 1 ; using formula log  (1 + x)/x   x tans to 0  =1  }

therefore d ( log x) /dx  =  1/x

hope might help you

=   1* 1/x = 1/x

• 14
find the maximum the value acosecx +bsinx ?
• -1
( x+1/x)12
• 0
Please see photo...Hit like if u like it

• 34
Sorry guys 3rd last line is lim h-0 log(1+x)/x=1.Make this change.  Everything else is correct
• 2
e ka power root x
• -6
Both integration and differentiation of êx is êx
• 1
Ex 4b q19
• 2
Root cosx (by first principle)
• 1
Derivate (e^sinx).cosx by 1st principle
• 2
Y=logx solveit N
• 1
Y=logx solveit
• 0
y = ln x
f(x) = ln x
f(x+h) = ln (x+h)
f'(x) =  [ln (x+h) - ln x] / h
= ln (x+h/x) / h
= 1/h * ln(1+h/x)
Since ln(1+h/x) -> h/x where h != 0,
f'(x) = 1/h * h/x = 1/x
• -1
Hi

• 4
Please find this answer

• 0
X2 sin x
• 0
Please find this answer

• 0
Please find this answer

• 0
Find differentiation

• 0
please I don't know
• 0
8capter 8.2exericise
• 0
X2 sin x.
• 1
So hard question
• 1
What are you looking for?