Differentiation Question 186
Question: If $ f(x)=\frac{1}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}+b^{2}}} $ , then $ {f}’(x) $ is equal to
[Kurukshetra CEE 1998]
Options:
A) $ \frac{x}{(a^{2}-b^{2})}[ \frac{1}{\sqrt{x^{2}+a^{2}}}-\frac{1}{\sqrt{x^{2}+b^{2}}} ] $
B) $ \frac{x}{(a^{2}+b^{2})}[ \frac{1}{\sqrt{x^{2}+a^{2}}}-\frac{2}{\sqrt{x^{2}+b^{2}}} ] $
C) $ \frac{x}{(a^{2}-b^{2})}[ \frac{1}{\sqrt{x^{2}+a^{2}}}+\frac{1}{\sqrt{x^{2}+b^{2}}} ] $
D) $ (a^{2}+b^{2})[ \frac{1}{\sqrt{x^{2}+a^{2}}}-\frac{2}{\sqrt{x^{2}+b^{2}}} ] $
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)=\frac{1}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}+b^{2}}} $
$ f(x)=\frac{1}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}+b^{2}}}.\frac{\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}+b^{2}}}{\sqrt{x^{2}+a^{2}}-\sqrt{x^{2}+b^{2}}} $
$ f(x)=\frac{1}{a^{2}-b^{2}}[ \sqrt{x^{2}+a^{2}}-\sqrt{x^{2}+b^{2}} ] $
$ f’(x)=\frac{1}{a^{2}-b^{2}}[ \frac{2x}{2\sqrt{x^{2}+a^{2}}}-\frac{2x}{2\sqrt{x^{2}+b^{2}}} ] $
$ f’(x)=\frac{x}{a^{2}-b^{2}}[ \frac{1}{\sqrt{x^{2}+a^{2}}}-\frac{1}{\sqrt{x^{2}+b^{2}}} ] $ .
BETA
