Let y = $(logx)^π + x^{logπ}$ ... (1) Now let $y_1$ = $(logx)^x$ and yz = $x^{logx}$ ⇒ y = $y_1+y_2$ ... (2) Differentiating (2) w.r.t. x, ${dy}/{dx}$ = ${dy_1}/{dx}+{dy_2}/{dx}$ ... (3) Now consider $y_1$ = $(logx)^π$ Taking log on both sides, log $y_1$ = x log (log x) Differentiating w.r.t. x, we get $1/y_1 {dy_1}/{dx}$ = x × $1/{logx}$ × $1/x$ + 1 - log (log x) ⇒ ${dy_1}/{dx}$ = $y_1(1/{logx} + log(logx))$ ⇒ ${dy_1}/{dx}$ = $log x^π (1/{logx} + log (logx))$ ... (4) ⇒ Now, consider $y_2$ = (log x) (log x) = $(logx)^2$ Differentiating w.r.t. x, we get $1/y_2 {dy_2}/{dx}$ = 2 log x × $1/x$ ⇒ ${dy_2}/{dx}$ = $y_2 ({2logx}/x)$ = $x^{logx} ({2logx}/x)$ ... (5) Using equations (3), (4) and (5), we get: ${dy}/{dx}$ = $logx^π(1/{logx}+log(logx))$ + $x^{logπ}({2logx}/x)$