Applications Of Derivatives Question 188
Question: If $ y=\frac{x}{2}\sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2}\log (x+\sqrt{x^{2}+a^{2}}) $ ,then $ \frac{dy}{dx}= $
[AISSE 1983]
Options:
A) $ \sqrt{x^{2}+a^{2}} $
B) $ \frac{1}{\sqrt{x^{2}+a^{2}}} $
C) $ 2\sqrt{x^{2}+a^{2}} $
D) $ \frac{2}{\sqrt{x^{2}+a^{2}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ y=\frac{x}{2}\sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2}\log (x+\sqrt{x^{2}+a^{2}}) $
therefore $ \frac{dy}{dx}=\frac{1}{2}[ \sqrt{a^{2}+x^{2}}+x\frac{1}{2}{{(a^{2}+x^{2})}^{-1/2}}2x ] $
$ +\frac{a^{2}}{2}\frac{1}{(x+\sqrt{(x^{2}+a^{2})}}[ 1+\frac{1}{2}{{(x^{2}+a^{2})}^{-1/2}}2x ] $
$ =\frac{n^{2}[({\sec^{n}\theta -\cos^{n}\theta})^{2}+4\sec^{n}\theta \cos^{n}\theta]}{({\sec \theta -\cos \theta})^{2}+4\sec \theta \cos \theta}=\frac{n^{2}(y^{2}+4)}{x^{2}+4} $
BETA
