Differentiation Question 6
Question 6 - 01 February - Shift 2
If $y(x)=x^{x}, x>0$, then $y^{\prime \prime}(2)-2 y^{\prime}(2)$ is equal to
(1) $8 \log _{e} 2-2$
(2) $4 \log _{e} 2+2$
(3) $4(\log _{e} 2)^{2}-2$
(4) $4(\log _{e} 2)^{2}+2$
Show Answer
Answer: (3)
Solution:
Formula: Successive differentiation, Differentiation of some elementary functions, Product Rule
$ \begin{aligned} & 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 n 2)^{2}+2 \\ & y^{\prime}(2)=4(1+\ell n 2) \\ & y^{\prime \prime}(2)-2 y^{\prime}(2)=4(1+\ell n 2)^{2}+2-8(1+\ell n 2) \\ & y^{\prime \prime}(2)-2 y^{\prime}(2) =4(1+\ell n 2)[1+\ell n 2-2]+2 \text{ } \\ & y^{\prime \prime}(2)-2 y^{\prime}(2) =4(\ell n 2)^{2}-1)+2 \\ & y^{\prime \prime}(2)-2 y^{\prime}(2) =4(\ell n 2)^{2}-2 \end{aligned} $
