Question

If $y(x)=x^x, x>0$, then $y^{\prime \prime}(2)-2 y^{\prime}(2)$ is equal to :

• A. $4\left(\log _{ e } 2\right)^2-2$
• B. $8 \log _{ e } 2-2$
• C. $4\left(\log _e 2\right)^2+2$
• D. $4 \log _{ e } 2+2$

Answer 1

Answer: (A)

$ y^{\prime}=x^x $
$ y^{\prime}=x^x(1+\ell n x) $
$ y^{\prime \prime}=x^x(1+\ell n x)^2+x^x \cdot \frac{1}{x} $
$ y^{\prime \prime}(2)=4(1+\ell \operatorname{n} 2)^2+2 $
$y^{\prime}(2)=4(1+\ell n 2) $
$ y^{\prime \prime}(2)-2 y^{\prime}(2)=4(1+\ell \operatorname{n} 2)^2+2-8(1+\ell n 2) $
$ =4(1+\ell n 2)[1+\ell n 2-2]+2 $
$ \left.=4(\ell n 2)^2-1\right)+2 $
$ =4(\ell n 2)^2-2$