Differentiation Question 8
Question: If $ y={{\sqrt{x}}^{{{\sqrt{x}}^{\sqrt{x}….\infty }}}} $ , then $ \frac{dy}{dx}= $
Options:
A) $ \frac{y^{2}}{2x-2y\log x} $
B) $ \frac{y^{2}}{2x+\log x} $
C) $ \frac{y^{2}}{2x+2y\log x} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ y={{\sqrt{x}}^{{{\sqrt{x}}^{\sqrt{x}…..\infty }}}}\Rightarrow y={{(\sqrt{x})}^{y}} $
Therefore $ \log y=y\log {x^{1/2}}=\frac{1}{2}y\log x $
Therefore $ \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}( \log x\frac{dy}{dx}+\frac{y}{x} )\Rightarrow \frac{dy}{dx}( \frac{2}{y}-\log x )=\frac{y}{x} $
Therefore $ \frac{dy}{dx}=\frac{y.y}{x(2-y\log x)}=\frac{y^{2}}{x(2-y\log x)} $ .
BETA
