Differentiation Question 209
Question: If $ y=x^{2}\log x+\frac{2}{\sqrt{x}}, $ then $ \frac{dy}{dx}= $
Options:
A) $ x+2x\log x-\frac{1}{\sqrt{x}} $
B) $ x+2x\log x-\frac{1}{{x^{3/2}}} $
C) $ x+2x\log x-\frac{2}{{x^{3/2}}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ y=x^{2}\log x+\frac{2}{\sqrt{x}} $
$ \frac{dy}{dx}=2x\log x+x-{x^{-3/2}}=x+2x\log x-\frac{1}{{x^{3/2}}} $ .
BETA
