In differentiation natural is represented by “ln”. ln = log e. Logarithmic differentiation is possible only when the base of the log is e. Otherwise we need to apply the change of base rule before going for differentiation.

Example log 10 x = log e (x )/ log e 10 = 1 / log e 10 `xx` lnx.

Now it can be differentitated.

The formula we need to remember are as follows:

(i) `d / dx` (log x) = `1 / x`

(ii) log (ab) = log a + log b.

(iii) log `(a / b)` = log a – log b.

(iv) log (a m) = m log a.

Here log means, we consider it as ln. That is log with base e.

Now let us see few problems on this topic differentiate natural log.

Example problems on differentitate natural log.

 

Ex 1: Differentiate: y = log x 2

Soln: Given: y = log x 2 = 2 log x

Therefore `dy / dx` = `d / dx` (2 log x) = 2 `d / dx` (log x) = `2 / x` .

Ex 2: Differentiate: y = log [(x + 2) (x + 3)]

Soln: Given: y = log [(x + 2) (x + 3)]

Therefore y = log (x + 2) + log (x + 3)

Therefore `dy / dx ` =` d / dx` [log (x + 2) + log (x + 3)]

`dy / dx ` = `d / dx` (log (x + 2)) + `d / dx` (log (x + 3))

= `1 /( x + 2)` + `1 /( x + 3)` =` (x + 3 + x + 2) /[ (x + 2) (x + 3)]`

=` (2x + 5) / (x ^2 + 5x + 6)`

Ex 3: Differentiate y =` log [(x + 3) /( x + 4)]`

Soln: Given: y = `log [(x + 3) / (x + 4)]`

`=>` y = log (x + 3) – log (x + 4)

Therefore `dy / dx ` = `d / dx ` (log (x + 3)) – `d / dx` log (x + 4)

= `1 / (x + 3) ** 1 /( x + 4)` = `[x + 4 ** (x + 3)] /[ (x + 3) (x + 4)]`

=`[ x + 4 ** x ** 3] /[ x ^2 + 7x + 12 ]` = `[2x + 1] / [x ^2 + 7x + 12]`

 

More example problems on differentitate natural log.

 

Ex 4: Differentiate y = log (e x)

Soln: Given: y = log (e x)

y = x . log e [Here log e ^e = 1]

`=>` y = x

Therefore `dy / dx ` = 1  [Since` d / dx` (x) = 1]

Ex 5: Differentiate y = x x

Soln: Since the variable in the power, we can apply log to the base e before going for differentiation.

Soln: Given: y = x x

Therefore log y = log x x = x . log x

Therefore on diff, `1 / y` . `dy / dx ` = x . `d / dx` (log x) + log x . `d / dx` (x)

`dy / dx` = y [x . `1 / x` + log x (1)] = y [1 + log x]

Therefore `dy / dx` = x x [ 1 + log x]