Question

Find `dy/dx,"if"  y=x^x+(logx)^x`

Answer 1

Let `y=x^x+(logx)^x`

Let u = x^x and v = `(logx)^x`       ...(1)

Then y = u + v

∴ `dy/dx= (du)/dx +(dv)/dx`

Take u = x^x

∴ log u = log x^x = x log x

Differentiating both sides w.r.t. x, we get

`1/u.(du)/dx =d/dx(xlogx)`

`=xd/dx(logx)+(logx).d/dx(x)`

`= x xx1/x+(logx)xx1`

∴ `(du)/dx =u(1+logx)=x^x(1+logx)`       ...(2) 

Also, v = (log x)^x

∴ log v = log (log x)^x = x log (log x)

Differentiating both sides w.r.t. x, we get

`1/v.(dv)/dx=d/dx[xlog(logx)]`

`=x.d/dx[log(logx)] + [log(logx)].d/dx(x)`

`=x xx1/logx .d/dx(logx)+[log(logx)]xx1`

`=x xx1/logx xx1/x+log(logx)`

∴ `(dv)/dx =v[1/logx+log(logx)]`

`=(logx)^x[1/logx+log(logx)]`        ...(3)

From (1), (2) and (3), we get

`dy/dx=x^x(1+logx)+(logx)^x[1/logx+log(logx)]`