Differentiation Question 6
Question: If $ y=x^{2}+{x^{\log x}}, $ then $ \frac{dy}{dx}= $
[DSSE 1986]
Options:
A) $ \frac{x^{2}+\log x.{x^{\log x}}}{x} $
B) $ x^{2}+\log x.{x^{\log x}} $
C) $ \frac{2(x^{2}+\log x.{x^{\log x}})}{x} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=x^{2}+{x^{\log x}} $
Therefore $ \frac{dy}{dx}=2x+{x^{\log x}}( 2{\log _{e}}x.\frac{1}{x} ) $
$ =\frac{2(x^{2}+{x^{\log x}}{\log _{e}}x)}{x} $ .
BETA
