Question

Differentiate (log x)x + x log x w.r.t. x

please explain step by step​

Answer 1

Step-by-step explanation:

y=(logx)

x

+x

logx

Let y=a+b

dx

dy

=?

a=(logx)

x

⇒loga=xlogx⇒

a

1

dx

da

=logx+

x

x

=1+lnx

⇒

dx

da

=a(1+lnx)=(logx)

x

(1+lnx)

b=x

logx

logb=(logx).x

b

1

.

dx

db

=

x

1

.x+logx

dx

db

=b(1+logx)=x

logx

(1+logx)

dx

dy

=

dx

da

+

dx

db

=(logx)

x

(1+logx)+x

logx

(1+logx)=(1+logx)((logx)

x

+x

logx

)

dx

dy

=(1+logx)y