Differentiation Question 395
Question: If $ y=x^{n}\log x+x{{(\log x)}^{n}} $ , then $ \frac{dy}{dx}= $
Options:
A) $ {x^{n-1}}(1+n\log x)+{{(\log x)}^{n-1}}[n+\log x] $
B) $ {x^{n-2}}(1+n\log x)+{{(\log x)}^{n-1}}[n+\log x] $
C) $ {x^{n-1}}(1+n\log x)+{{(\log x)}^{n-1}}[n-\log x] $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y=x^{n}\log x+x{{(\log x)}^{n}} $
$ \frac{dy}{dx}=n{x^{n-1}}\log x+x^{n}.( \frac{1}{x} )+xn{{(\log x)}^{n-1}}.( \frac{1}{x} )+1.{{(\log x)}^{n}} $
$ ={x^{n-1}}(1+n\log x)+{{(\log x)}^{n-1}}[n+\log x] $ .
BETA
