Differentiation Question 38
Question: If $ y={x^{(x^{x})}} $ , then $ \frac{dy}{dx} $ is
Options:
A) $ y[ x^{x}(logex)logx+x^{x} ] $
B) $ y[ x^{x}(logex)logx+x ] $
C) $ y[ x^{x}(logex)logx+{x^{x-1}} ] $
D) $ y[ x^{x}(log _{e}x)logx+{x^{x-1}} ] $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ y={x^{(x^{x})}} $ Or $ \log y=x^{x}\log x $ Or $ \frac{1}{y}\frac{dy}{dx}=\frac{dz}{dx}\log x+\frac{1}{x}z $
(where $ x^{x}=z $ ) Or $ \frac{dy}{dx}={x^{(x^{x})}}[ x^{x}(log _{e}x)logx+{x^{x-1}} ] $
$ ( \therefore \frac{dz}{dx}=x^{x}{\log _{e}}x ) $
BETA
