Applications Of Derivatives Question 197
Question: If $ y=f( \frac{2x-1}{x^{2}+1} ) $ and $ {f}’(x)=\sin x^{2}, $ then $ \frac{dy}{dx}= $
[IIT 1982]
Options:
A) $ \frac{6x^{2}-2x+2}{{{(x^{2}+1)}^{2}}}\sin {{( \frac{2x-1}{x^{2}+1} )}^{2}} $
B) $ \frac{6x^{2}-2x+2}{{{(x^{2}+1)}^{2}}}{{\sin }^{2}}( \frac{2x-1}{x^{2}+1} ) $
C) $ \frac{-2x^{2}+2x+2}{{{(x^{2}+1)}^{2}}}{{\sin }^{2}}( \frac{2x-1}{x^{2}+1} ) $
D) $ \frac{-2x^{2}+2x+2}{{{(x^{2}+1)}^{2}}}\sin {{( \frac{2x-1}{x^{2}+1} )}^{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ t=\frac{2x-1}{x^{2}+1}, $ then $ \frac{dy}{dx}=f’(t).\frac{dt}{dx} $
$ =\sin t^{2}\frac{d}{dx}( \frac{2x-1}{x^{2}+1} )=\frac{2(1+x-x^{2})}{{{(1+x^{2})}^{2}}}.\sin {{( \frac{2x-1}{x^{2}+1} )}^{2}} $ .
BETA
