Question

find dy/dx of y=x^(x) + x^(logx)​

Answer 1

Actually welcome to the concept of the Derivatives

•Here, basically we are going to the use the Chain rule, logarithm and exponential form

•So after solving we get as

•Applying the logrithms on the both sides we get as

∆dy/dx = x^x(1+logx) + 2x^logx(logx/x)

Answer 2

${\mathcal{\red{SOLUTION}}}$

$\tt{y = (\log\;x)^{x}+x^{\log\;x}}$

$\tt{\implies y = e^{\log\;(\log\;x)^{x}} + e^{\log\;(x^{\log\;x})}}$

$\tt{\implies y = e^{x\;\log\;(\log\;x)} + e^{\log x.\log x}}$

$\tt{\implies \dfrac{dy}{dx} = e^{x \log (\log x)}\times \dfrac{d}{dx} [x \log (\log x)]+ e^{(\log x)^{2}} \times \dfrac{d}{dx} [{\log x)^{2}}]}$

$\tt{\implies \frac{dy}{dx} = (\log x)^{x} \bigg[x \times \dfrac{1}{\log x} \times \dfrac{1}{x}+\log (\log x)\bigg]+x^{\log x} \times \dfrac{2 \log x}{x}}$

$\tt{\implies \dfrac{dy}{dx} = (\log x)^{x} \bigg[\log (\log x)+\dfrac{1}{\log\;x}\bigg] + x^{\log x} \times \dfrac{2\log x}{x}}$