JEE Main 2020DifferentiationDifferentiationMediumQuestionIf y(x)=xx,x>0y(x)=x^{x},x > 0y(x)=xx,x>0, then y′′(2)−2y′(2)y''(2)-2y'(2)y′′(2)−2y′(2) is equal toOptionsA4(loge2)2+24(\log_{e}2)^{2}+24(loge2)2+2B8loge2−28\log_{e}2-28loge2−2C4loge2+24\log_{e}2+24loge2+2D4(loge2)2−24(\log_{e}2)^{2}-24(loge2)2−2Check AnswerHide SolutionSolutiony=xxy′=xx(1+lnx)y′′=xx(1+lnx)2+xxxf′′(2)−2f′(2)=(4(1+ln2)2+2)−(2)(4(1+ln2))=4(1+(ln2)2)+2−8=4(ln2)2−2\begin{aligned} & y=x^x \\\\ & y^{\prime}=x^x(1+\ln x) \\\\ & y^{\prime \prime}=x^x(1+\ln x)^2+\frac{x^x}{x} \\\\ & f^{\prime \prime}(2)-2 f^{\prime}(2)=\left(4(1+\ln 2)^2+2\right)-(2)(4(1+\ln 2)) \\\\ & =4\left(1+(\ln 2)^2\right)+2-8 \\\\ & =4(\ln 2)^2-2 \\\\ & \end{aligned}y=xxy′=xx(1+lnx)y′′=xx(1+lnx)2+xxxf′′(2)−2f′(2)=(4(1+ln2)2+2)−(2)(4(1+ln2))=4(1+(ln2)2)+2−8=4(ln2)2−2