Differentiation Question 458
Question: If $ y={x^{(x^{x})}} $ , then $ \frac{dy}{dx}= $
[AISSE 1989]
Options:
A) $ y[x^{x}(\log ex).\log x+x^{x}] $
B) $ y[x^{x}(\log ex).\log x+x] $
C) $ y[x^{x}(\log ex).\log x+{x^{x-1}}] $
D) $ y[x^{x}({\log _{e}}x).\log x+{x^{x-1}}] $
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Answer:
Correct Answer: C
Solution:
$ y={x^{(x^{x})}}\Rightarrow \log y=x^{x}\log x $
Therefore $ \frac{1}{y}\frac{dy}{dx}=\frac{dz}{dx}.\log x+\frac{1}{x}.z $ , (where $ x^{x}=z $ )
$ \Rightarrow \frac{dy}{dx}={x^{(x^{x})}}[ x^{x}(\log ex).\log x+{x^{x-1}} ] $ , $ { \because \frac{dz}{dx}=x^{x}\log ex } $ .
BETA
