Differentiation Question 446
Question: If $ y={x^{\sqrt{x}}}, $ then $ \frac{dy}{dx} $ =
Options:
A) $ {x^{\sqrt{x}}}\frac{2+\log x}{2\sqrt{x}} $
B) $ {x^{\sqrt{x}}}\frac{2+\log x}{\sqrt{x}} $
C) $ \frac{2+\log x}{2\sqrt{x}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={x^{\sqrt{x}}}\Rightarrow {\log _{e}}y=\sqrt{x}\log x $
Therefore $ \frac{1}{y}\frac{dy}{dx}=\sqrt{x}\frac{1}{x}+\frac{1}{2\sqrt{x}}\log x $ or $ \frac{dy}{dx}={x^{\sqrt{x}}}[ \frac{2+{\log _{e}}x}{2\sqrt{x}} ] $
BETA
