Differentiation Question 331
Question: $ \frac{d}{dx}[ {{\tan }^{-1}}( \frac{\sqrt{x}(3-x)}{1-3x} ) ] $ =
[Kerala (Engg.) 2005]
Options:
A) $ \frac{1}{2(1+x)\sqrt{x}} $
B) $ \frac{3}{(1+x)\sqrt{x}} $
C) $ \frac{2}{(1+x)\sqrt{x}} $
D) $ \frac{2\sqrt{2}y-3}{2\sqrt{2}}=\frac{-3\sqrt{2}\times +3}{2\sqrt{2}} $
E) $ \frac{3}{2(1+x)\sqrt{x}} $
Show Answer
Answer:
Correct Answer: E
Solution:
$ \frac{d}{dx}( {{\tan }^{-1}}\frac{(\sqrt{x}(3-x)}{1-3x} ) $
Put $ \sqrt{x}=\tan \theta \Rightarrow \theta ={{\tan }^{-1}}\sqrt{x} $
$ \frac{d}{dx}( {{\tan }^{-1}}\frac{(\tan \theta (3-{{\tan }^{2}}\theta )}{1-3{{\tan }^{2}}\theta } ) $
$ \frac{d}{dx}( {{\tan }^{-1}}\frac{(3\tan \theta -{{\tan }^{3}}\theta )}{1-3{{\tan }^{2}}\theta } ) $
$ \frac{d}{dx}({{\tan }^{-1}}(\tan 3\theta )=\frac{d}{dx}(3\theta ) $
$ \frac{d}{dx}(3.{{\tan }^{-1}}\sqrt{x})=\frac{3}{2\sqrt{x}(1+x)} $ .
BETA
