Differentiation Question 311
Question: If $ x=\exp { {{\tan }^{-1}}( \frac{y-x^{2}}{x^{2}} ) } $ , then $ \frac{dy}{dx} $ equals
[MP PET 2002]
Options:
A) $ 2x[1+\tan (\log x)]+x{{\sec }^{2}}(\log x) $
B) $ x[1+\tan (\log x)]+{{\sec }^{2}}(\log x) $
C) $ 2x[1+\tan (\log x)]+x^{2}{{\sec }^{2}}(\log x) $
D) $ 2x[1+\tan (\log x)]+{{\sec }^{2}}(\log x) $
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Answer:
Correct Answer: A
Solution:
$ x=\exp { {{\tan }^{-1}}( \frac{y-x^{2}}{x^{2}} ) } $
Therefore $ \log x={{\tan }^{-1}}( \frac{y-x^{2}}{x^{2}} ) $
Therefore $ \frac{y-x^{2}}{x^{2}}=\tan (\log x) $
Therefore $ y=x^{2}\tan (\log x)+x^{2} $
Therefore $ \frac{dy}{dx}=2x.\tan (\log x)+x^{2}.\frac{{{\sec }^{2}}(\log x)}{x}+2x $
Therefore $ \frac{dy}{dx}=2x\tan (\log x)+x{{\sec }^{2}}(\log x)+2x $
Therefore $ \frac{dy}{dx}=2x[1+\tan (\log x)]+x{{\sec }^{2}}(\log x) $ .
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