Question

If y = (log x)^x + x^log x, find `"dy"/"dx".`

Answer 1

Let y =(log x)^x + x^log x

Also, let u =(log x)^x and  v = x^log x

∴ y = u + v

`⇒"dy"/"dx" = "du"/"dx"+"dv"/"dx"` ........(1)

u = (logx)^x

⇒ log u = log[(log x)^x]

⇒ log u = x log(log x)

Differentiating both sides with respect to x, we obtain

`1/"u" "du"/"dx" ="d"/"dx"("x") xx log(log"x")+"x"."d"/"dx"[log(log"x")]`

`⇒"du"/"dx" = "u"[1xxlog(log"x")+"x". 1/log"x"."d"/"dx"(log"x")]`

`⇒"du"/"dx"=(log"x")^"x"[log(log"x")+"x"/(log"x"). 1/"x"]`

`⇒"du"/"dx"=(log"x")^"x"[log(log"x")+1/(log"x")]`

`⇒"du"/"dx"=(log"x")^"x"[(log(log"x").log"x"+1)/log"x"]`

`⇒"du"/"dx"=(log"x")^("x"-1)[1+log"x".log(log"x")]` .....(2)

v = x^logx

⇒ log v = log(x^logx)

⇒ log v = log  x log x = (log x)²

Differentiating both sides with respect to x, we obtain

`1/"v"."dv"/"dx"="d"/"dx"[(log"x")^2]`

`⇒ 1/"v"."dv"/"dx"=2(log"x")."d"/"dx"(log"x")`

`⇒"dv"/"dx" = 2"v"(log"x"). 1/"x"`

`⇒"dv"/"dx" = 2"x"^(log"x") log"x"/"x"`

`⇒"dv"/"dx"=2"x"^(log"x"-1) . log"x"` ......(3)

Therefore, from (1), (2), and (3), we obtain

`"dy"/"dx" = (log"x")^("x"-1)[1+log"x".log(log"x")]+2"x"^(log"x"-1) .log"x"`