Question

(x-1) (x-2)Vix-3) (x-4) (x-5)2.-·

Answer 1

Given expression is ( x−1 )( x−2 ) ( x−3 )( x−4 )( x−5 ) .

Let the expression be,

y= ( x−1 )( x−2 ) ( x−3 )( x−4 )( x−5 ) y= ( ( x−1 )( x−2 ) ( x−3 )( x−4 )( x−5 ) ) 1 2

Take log on both sides of the given expression.

logy=log ( ( x−1 )( x−2 ) ( x−3 )( x−4 )( x−5 ) ) 1 2 logy= 1 2 log( ( x−1 )( x−2 ) ( x−3 )( x−4 )( x−5 ) )

As we know that, log( a b )=bloga . logy= 1 2 [ log( ( x−1 )( x−2 ) )−log( ( x−3 )( x−4 )( x−5 ) ) ] logy= 1 2 ( log( x−1 )+log( x−2 )+log( x−3 )+log( x−4 )+log( x−5 ) )

d( logy ) dx = 1 2 d( log( x−1 )+log( x−2 )+log( x−3 )+log( x−4 )+log( x−5 ) ) dx d( logy ) dx ( dy dy )= d( log( x−1 ) ) dx + d( log( x−2 ) ) dx + d( log( x−3 ) ) dx + d( log( x−4 ) ) dx + d( log( x−5 ) ) dx 1 y ( dy dx )= 1 2 ( 1 x−1 + 1 x−2 + 1 x−3 + 1 x−4 + 1 x−5 ) dy dx = 1 2 y( 1 x−1 + 1 x−2 + 1 x−3 + 1 x−4 + 1 x−5 )

Now, differentiating the expression on both sides with respect to x, we get

Now, substitute y= ( x−1 )( x−2 ) ( x−3 )( x−4 )( x−5 ) in the above equation.

dy dx = 1 2 ( x−1 )( x−2 ) ( x−3 )( x−4 )( x−5 ) ( 1 x−1 + 1 x−2 + 1 x−3 + 1 x−4 + 1 x−5 )