Question

Differentiate the following w.r.t. x:

`x^(x^x) + e^(x^x)`

Answer 1

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

Put u = `x^(x^x) and v = e^(x^(x)`

Then y = u + v

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

Take u = `x^(x^(x)`

∴ log u = `logx^(x^(x)) = x^x*logx`

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

`1/u*"du"/"dx" = "d"/"dx"(x^x*logx)`

= `x^x*"d"/"dx"(logx) + (logx)*"d"/"dx"(x^x)`

= `x^x xx 1/x + (logx)*"d"/"dx"(x^x)`      ...(2)

To find `"d"/"dx"(x^x)`

Let ω = x^x

Then log ω = xlogx

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

`1/omega*"dω"/"dx" = "d"/"dx"(xlogx)`

= `x*"d"/"dx"(logx) + (logx)*"d"/"dx"(x)`

= `x xx (1)/x + (logx) xx 1`

∴ `"dω"/"dx" = omega(1 + logx)`

∴ `"d"/"dx"(x^x) = x^x(1 + logx)`      ...(3)

∴ from (2),

`1/u*"du"/"dx" = x^x xx (1)/x + (logx)*x^x(1 + logx)`

∴ `"du"/"dx" = y[x^x xx 1/x + (logx)*x^x(1 + logx)]`

= `x^(x^x)*x^x[1/x + (logx)*(1 + logx)]`

= `x^(x^x)*x^x*logx[1 + logx + 1/(xlogx)]`      ...(4)

Also, v = `e^(x^(x)`

∴ `"dv"/"dx" = "d"/"dx"(e^(x^x))`

= `e^(x^(x))*"d"/"dx"(e^(x^x))`

= `e^(x^(x))*x^x(1 + logx)`      ...(5) [By (3)] 

From (1), (4) and (5), we get              

`"dy"/"dx" = x^(x^x)*x^x*logx[1 + logx + 1/(xlogx)] + e^(x^x)*x^x(1 + logx)`