Question

Differentiate the following w.r.t. x : `10^(x^(x)) + x^(x(10)) + x^(10x)`

Answer 1

Let y = `10^(x^(x)) + x^(x(10)) + x^(10x)`
Put u = `10^(x^(x)), v = x^(x^(10)) and omega = x^(10^(x)`
Then y = u + v + ω
∴ `"dy"/"dx" = "u"/"dx" + "dv"/"dx" + "dω"/"dx"`                   ...(1)
Take, u = `10^(x^(x)`
∴ `"du"/"dx" = "d"/"dx"(10^(x^(x)`

= `10^(x^(x)).log10."d"/"dx"(x^x)`

To find `"d"/"dx"(x^x)`
Let z = x^x
∴ logz = logx^x = xlogx
Differentiating both sides w.r.t. x, we get
`1/z."dz"/"dx" = "d"/"dx"(xlogx)`

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

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

∴ `"dz"/"dx" = z(1 + logx)`

∴ `"d"/"dx"(x^x) = x^x(1 + logx)`

∴ `"du"/"dx" = 10^(x^x).log10.x^x(1 + logx)`  ...(2)
Take, v = `x^(x^10)`
∴ log v = `logx^(x^10) = x^10.logx`
Differentiating both sides w.r.t. x, we get
`1/v."dv"/"dx" = "d"/"dx"(x^10logx)`

= `x^10."d"/"dx"(logx) + (logx)."d"/"dx"(x^10)`

= `x^10 xx 1/x + (logx)(10x^9)`

∴ `"dv"/"dx" = v[x^9 + 10x^9logx]`

∴ `"dv"/"dx" = x^(x^10).x^9(1 + 10logx)`       ...(3)
Also, ω = `x^(10x)`
∴ log ω = `logx^(10x) = 10^x.logx`
Differentiating both sides w.r.t. x, we get
`1/omega ."dω"/"dx" = "d"/"dx"(10^x.logx)`

= `10^x."d"/"dx"(logx) + (logx)."d"/"dx"(10^x)`

= `10^x xx 1/x + (logx)(10^x.log10)`

∴ `"dω"/"dx" = ω[10^x/x + 10^x.(logx)(log10)]`

∴ `"dω"/"dx" = x^(10x).10^x[1/x + (logx)(log10)]` ...(4)
From (1),(2),(3) and (4), we get
`"dy"/"dx" = 10^(x*x).log10.x^x(1 + logx) + x^(x^10).x^9(1 + 10logx) + x^(10x).10^x[1/x + (logx)(log10)]`.