Differentiation Question 448
Question: If $ y={{(\sin x)}^{{{(\sin x)}^{(\sin x)……\infty }}}} $ , then $ \frac{dy}{dx}= $
Options:
A) $ \frac{y^{2}\cot x}{1-y\log \sin x} $
B) $ \frac{y^{2}\cot x}{1+y\log \sin x} $
C) $ \frac{y\cot x}{1-y\log \sin x} $
D) $ \frac{y\cot x}{1+y\log \sin x} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={{(\sin x)}^{{{(\sin x)}^{(\sin x)…..\infty }}}} $
Therefore $ y={{(\sin x)}^{y}}\Rightarrow {\log _{e}}y=y\log \sin x $
Therefore $ \frac{1}{y}\frac{dy}{dx}=\frac{dy}{dx}[\log \sin x+y\cot x] $
$ \therefore \frac{dy}{dx}=\frac{y^{2}\cot x}{1-y\log \sin x} $ .
BETA
