MCQ
If $y = {({x^x})^x}$, then ${{dy} \over {dx}} =$
- A${({x^x})^x}(1 + 2\log x)$
- B${({x^x})^x}(1 + \log x)$
- ✓$x{({x^x})^x}(1 + 2\log x)$
- D$x\,{({x^x})^x}(1 + \log x)$
✓
Answer
Correct option: C.
$x{({x^x})^x}(1 + 2\log x)$
c
(c) $y = {({x^x})^x} \Rightarrow {\log _e}y = x{\log _e}{(x)^x} = {x^2}.{\log _e}x$
(c) $y = {({x^x})^x} \Rightarrow {\log _e}y = x{\log _e}{(x)^x} = {x^2}.{\log _e}x$
==> $\frac{1}{y}\frac{{dy}}{{dx}} = {x^2}.\frac{1}{x} + 2x.{\log _e}x$
$\therefore \frac{{dy}}{{dx}} = x{({x^x})^x}[1 + 2{\log _e}x]$.
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