Dy/dx of

Question

y=(lnx)^(x^2).

Answer 1

Given function : y = (lnx)^x²

Apply log each side with base e.

lny = x²ln (lnx)

Apply derivative with respect to x

(d/dx)(lny) = (d/dx)(x²ln(lnx))

Apply formulas : (d/dx)(UV)= V(dU/dx)+U(dV/dx) and (d/dx)(lnU) = (1/U)(dU/dx)

Here consider U = x² and V = ln (lnx)

(1/y)(dy/dx) =(ln(lnx))(d/dx)(x²)+(x²)(d/dx)(ln(lnx))

Apply formulas : (d/dx)(f)ⁿ = (nf^n-1)(df/dx)  and   (d/dx)(lnU) = (1/U)(dU/dx)

(1/y)(dy/dx) =(ln(lnx))(2x)+(x²)(1/lnx) (d/dx)(lnx)

Apply formula : (d/dx)(lnx) = 1/x

(1/y)(dy/dx) =2xln(lnx)+(x²)(1/lnx) (1/x)

(dy/dx) =y [ 2xln(lnx)+(x)(1/lnx)]

Substitute y = (lnx)^x²

(dy/dx) =(lnx)^x²[ 2xln(lnx)+(x/lnx) ]

The solution is  (dy/dx) =(lnx)^x²[ 2xln(lnx)+(x/lnx) ].