Differentiation Question 219
Question: If $ y=\frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}} $ , then $ \frac{dy}{dx}= $
[AISSE 1986]
Options:
A) $ \frac{ay}{x\sqrt{a^{2}-x^{2}}} $
B) $ \frac{ay}{\sqrt{a^{2}-x^{2}}} $
C) $ \frac{ay}{x\sqrt{x^{2}-a^{2}}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y=\frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}\Rightarrow y=\frac{{{(\sqrt{a+x}-\sqrt{a-x})}^{2}}}{(a+x)-(a-x)} $
$ \Rightarrow y=\frac{(a+x)+(a-x)-2(\sqrt{a^{2}-x^{2}})}{2x} $
$ =\frac{2a-2\sqrt{a^{2}-x^{2}}}{2x} $ or $ y=\frac{a-\sqrt{a^{2}-x^{2}}}{x} $ ……..(i) Differentiating w.r.t. x of y, we get $ \frac{dy}{dx}=\frac{x[ -\frac{1}{2\sqrt{a^{2}-x^{2}}}(-2x) ]-(a-\sqrt{a^{2}-x^{2}})}{x^{2}} $
$ =\frac{x^{2}-a\sqrt{a^{2}-x^{2}}+a^{2}-x^{2}}{x^{2}\sqrt{a^{2}-x^{2}}}=\frac{a(a-\sqrt{a^{2}-x^{2}})}{x^{2}\sqrt{a^{2}-x^{2}}} $
$ =\frac{a}{x\sqrt{a^{2}-x^{2}}}[ \frac{a-\sqrt{a^{2}-x^{2}}}{x} ]=\frac{ay}{x\sqrt{a^{2}-x^{2}}} $ [By (i)]
BETA
