Question

derivative of 1)logx by first principle

Answer 1

Hey mate!!!

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        

HOPE IT HELPS.

Answer 2

Hello Friend..❤️❤️

The answer of u r question is..✌️✌️

Derivate of logx by first principal.


Ans:✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️✍️


Given,



$f(x) = logx$


$f(x + h) = log(x + h)$


$\frac{dy}{dx} = f(x + h) - f \frac{(x)}{h}$



$= log(x + h) - logx$


$= log \frac{(x + h)}{x}$


$= log \frac{(1 + \frac{h}{x}) }{h}$


$= \frac{d(logx)}{dx} = \frac{1}{x}$



$= \frac{1 \times 1}{x} = \frac{1}{x}$


Thank you...⭐️⭐️⭐️