Ts Eamcet
Differentiation
TG EAPCET 2025 (Online) 4th May Morning Shift
MCQ+1 / -02025
If $y=x^{\log x}+(\log x)^x, x>1$, then $\left(\frac{d y}{d x}\right)_{x=e}=$
Ts Eamcet
TG EAPCET 2025 (Online) 4th May Morning Shift
If $y=x^{\log x}+(\log x)^x, x>1$, then $\left(\frac{d y}{d x}\right)_{x=e}=$