Question

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

Answer 1

Let us suppose y = u + v

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

where `u = [log (underlinex)]^x`

And v = x^logx   ...(ii)

First take u = [log(x)]^x

Taking log of both the sides

logu = xloglogx   ...(iii)

Now differentiating the above w.r. to x

⇒ `1/"u" * "du"/"dx" = x * 1/logx * 1/x + log log x`

⇒ `"du"/"dx" = "u"(1/logx + log log x)`

⇒ `"du"/"dx" - (log(x))^x (1/logx + log log x)`   ...(iv)

Now take v = x^logx

log v = logx.logx   ...[Taking log on both sides]

= (log x)²

Differentiating w.r.t. x

⇒ `1/v * "dv"/"dx" = 2logx * 1/x`

⇒ `"dv"/"dx" = "v"(2log x * 1/x)`

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

⇒ `"dv"/"dx" = 2x^(logx - 1) * logx`   ...(v)

From equations (i), (iv) and (v)

`"dy"/"dx" = (log x)^x (1/log x + log log x) + 2x^(logx - 1) * log x`