Let y = (logx)π+xlogπ ... (1)
Now let y1 = (logx)x and yz = xlogx
⇒ y = y1+y2 ... (2)
Differentiating (2) w.r.t. x,

dy

dx

=

dy1

dx

+

dy2

dx

... (3)
Now consider y1 = (logx)π
Taking log on both sides,
log y1 = x log (log x)
Differentiating w.r.t. x, we get

1

y1

dy1

dx

= x ×

1

logx

×

1

x

+ 1 - log (log x)
⇒

dy1

dx

= y1(

1

logx

+log(logx))
⇒

dy1

dx

= logxπ(

1

logx

+log(logx)) ... (4)
⇒ Now, consider y2 = (log x) (log x) = (logx)2
Differentiating w.r.t. x, we get

1

y2

dy2

dx

= 2 log x ×

1

x

⇒

dy2

dx

= y2(

2logx

x

) = xlogx(

2logx

x

) ... (5)
Using equations (3), (4) and (5), we get:

dy

dx

= logxπ(

1

logx

+log(logx)) + xlogπ(

2logx

x

)