Question

find dy÷dx of x2logxsinx

Answer 1

$y = {x}^{2} log_{}(x) \sin(x) \\ dy \div dx = 2x( log(x) \sin(x) ) + {x}^{2} (1 \div x) \sin(x) + {x}^{2} log(x) \cos( x) \\ dy \div dx = 2x log(x) \sin(x) + x \sin(x) + {x}^{2} log(x) \cos(x)$
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Answer 2

Answer:

y' =  x ( x . log x . cos x + 2 . log x . sin x + sin x )

Step-by-step explanation:

Given :

y = x². log x. sin x

We have find its derivative :

We know product rule for three function :

i.e.  ( u . v . w )' = ( u . v ) ( w )' + ( v . w ) u' + ( u . w ) v'

= > d / d x ( sin x ) = cos x

= > d / d x ( log x ) = 1 / x

y = x². log x. sin x

= > d y / d x = ( x² . log x ) ( sin x )' + ( log x . sin x ) ( x² )' + ( x² . sin x ) ( log x )'

= > d y /d x = x² . log x . cos x + log x . sin x . 2 x + x² . sin x . 1 / x

= >  d y /d x = x² . log x . cos x + 2 x . log x . sin x + x . sin x

= > d y  / d x = x ( x . log x . cos x + 2 . log x . sin x + sin x )

Hence we get required answer.