Introduction of Derivative of a Function:

If a variable x changes from one value to another, the difference of the two values is called an increment in x. This increment may be positive or negative accordingly as x increases or decreases.

Let a particle moves along a path y = f(x). It moves `delx`

in x direction and the corresponding increase is `dely` in y. Then

`del y = f(x+delx) - f(x)`

`(dely)/(delx) =( (f(x+delx)-f(x)))/(delx)`

In `(dely)/(delx), delx` is very small distance moved by particle so it tends to zero.

 

The limit of `(dely)/(delx)` as `delx rarr0` , if it exists , it is called the derivative or differential coefficieint of y with respect to x and is denoted by `(dy)/(dx)`

Then we say `(dy)/(dx) = lim_(delx rarr 0) (dely)/(delx)`

`rArr (dy)/(dx) = lim_(delxrarr0) ( f(x+delx) -f(x))/(del x)`

If the increment `delx` is denoted by h , then the above formula becomes

`(dy)/(dx) = lim_(hrarr0) ( f(x+h)-f(x))/h` .

It is known as derivative or differential coefficient of f(x) with respect to x. We write `(dy)/(dx) ` or y' or y1.

 

Derivative of log base 10

 

      We use the definition of derivatives to find the derivative of log base 10.

 Let f(x) = log10 x

First we use the change of base to logarithms formula

Using the property of log:-      logb = `(logb)/(logf)" x "(logf)/(logc)` 

                                                          = (logb).(logf)                             here b,c,f is constant

    so    f(x) = loge x . log10 e

   Now log10 e is a constant. Therefore,

                    `(d)/(dx)(log_10 x) = (d)/(dx)(log_10 e . log_e x)`

                                               =` log_10 e . (d)/(dx) (log_e x) .............................................(i)`

First finding the value of `(d)/(dx)(log_e x) `

and put in above equation (i)

Let  g(x) = logx

 then g(x + h) = loge (x + h)

`(d)/(dx)(log_e x) = lim_(h->0)(log_e (x + h) - log_e x)/(h)`

                          = `lim_(h->0)(log(x + h)/h)/(h)`

                         = `lim_(h->0)" " (1)/(h) log (1 + h/x)`

                           = `lim_(h->0) " "(1)/(h)[(h)/(x)" - "(1)/(2)((h/x)^2)" + "(1)/(3)((h/x)^3) " - " .... ]`

                            = `lim_(h->0) ((1)/(x)" - " (1)/(2).(h)/(x^2)" + " (1)/(3).(h^2)/(x^3)" - " ........)`

                              = `(1)/(x)`

`(d)/(dx)(log_e x) = (1)/(x)`

putting the value in equation (i) then 

`(d)/(dx) log_10 x " = " (log_10 e). (1)/(x)`

 `(d)/(dx)(log_10 x) " = " (1)/(x) log_10 e`

so we use directly use the formula to find  the derivative of log base 10

Formula to find Derivative log base e

 `(d)/(dx)(log_e x) = (1)/(x)`

Lets use this formula to solve examples.

 

Examples based on derivative of log base 10 :

 

Ex 1: Find the Derivative of loge (log10x) with respect to x. 

Sol:  `(d)/(dx) log_e(log_10 x) = (1)/(log_10 x). (d)/(dx)(log_10 x)`

                                                = `(1)/(log_10 x)((1)/(x)log_10 e)`

                                                = `(1)/(x. log_10 x )*(log_10 e)`

Ex 2 Find the Derivative of log10 `(1)/(x)`  with respect to x. 

Sol:    `(d)/(dx) (log_10 (1)/(x)) = (1)/((1)/(x)). log_10 e`

                                                 = x log10 e