Differentiation Question 444
Question: If $ y={{(1+x)}^{x}}, $ then $ \frac{dy}{dx}= $
Options:
A) $ {{(1+x)}^{x}}[ \frac{x}{1+x}+\log ex ] $
B) $ \frac{x}{1+x}+\log (1+x) $
C) $ {{(1+x)}^{x}}[ \frac{x}{1+x}+\log (1+x) ] $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ y={{(1+x)}^{x}} $
Taking log on both sides, $ \log y=x\log (1+x) $
Differentiating w.r.t. x, we get $ \frac{1}{y}\frac{dy}{dx}=\log (1+x)+x\frac{1}{(1+x)} $
Thus $ \frac{dy}{dx}={{(1+x)}^{x}}[ \frac{x}{1+x}+\log (1+x) ] $
BETA
