Using differentials, find the approximate value of each of the following up to 3 places of decimal. (v) (0.999)^1/10 (vi) (15)^1/4

(v)  

Let  ……….(i)

   = 

     ……….(ii)

Now, from eq. (i), 

=  =  ……….(iii)

Here  and 

Then 

= 

  = 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),

Therefore, approximate value of  is 1 – 0.0001 = 0.9999.

(vi) 

Let  ……….(i)

   = 

   ……….(ii)

Now, from eq. (i), 

= =  ……….(iii)

Here  and 

Then 

= 

  = 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),  

Therefore, approximate value of  is  = 1.96875.