12.. DQiuffaedrernattiiactEioqnuations                                                                       eeLLeeaarrnn..PPuunnjjaabb
                                                                                                                version: 1.1
                =     1  loga x                                          lim  (1 +      1  =e 
                      x
                                                                             z→0        z)z

	                                		              		

   or  d   loga=x      1. 1                                        l=ogae        lo=1gea  1
       dx                  x ln a                                                                        
	                                                                                                ln    a  

                       ( )Find dy=if y
                             dx
Example 1:                                          log10  ax2 + bx + c

             	

Solution: Let u = ax2 + bx + c Then

		 y=              log10u  ⇒ dy=                 1        1
                              du                 u      In 10

      and du = d (ax + bx + c)= a(2x) + b(1)= 2ax + b

	 dx dx

   Th=us dy           d=y . du            1  .    1      du
          dx          du dx               u     ln 10    dx
	                                                      

                                  1                   (2ax + b)
                         ax2 + bx + c
=                                                ln 10

    ( )		

( )	    d                                                          2ax + b
        dx                                                        + bx + c)
	
   or       log10      ax2 + bx + c                 =      2             ln  10
                                                          (ax

Example 2:             ( )Differentiate ln x2 + 2x w.r.t. ' x '.

             	

( )Solution: 	 Let =y ln x2 + 2x , then

   d=y     d                                           1          .d
   dx      dx                                       x2 + 2x        dx
( ) ( ) ( )	                                                                            (Using chain rule)
                ln  x2 + 2x=                                       x2 + 2x

   =       x2   1  2x    .(=2x + 2)           2( x +1)
                +
( )	                                          x2 + 2x
        Thus
               d                            =2x(2x++21x)
	              dx
                   ln  x2 + 2x

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