Differentiation Question 39
Question: If $ y={{( x+\sqrt{x^{2}+a^{2}} )}^{n}} $ , then $ \frac{dy}{dx} $ is
Options:
A) $ \frac{ny}{\sqrt{x^{2}+a^{2}}} $
B) $ -\frac{ny}{\sqrt{x^{2}+a^{2}}} $
C) $ \frac{nx}{\sqrt{x^{2}+a^{2}}} $
D) $ -\frac{nx}{\sqrt{x^{2}+a^{2}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \frac{dy}{dx}=\frac{d}{dx}[ {{(x+\sqrt{x^{2}+a^{2}})}^{n}} ] $
$ =n{{( x+\sqrt{x^{2}+a^{2}} )}^{n-1}}.\frac{d}{dx}( x+\sqrt{x^{2}+a^{2}} ) $
$ =n{{( x+\sqrt{x^{2}+a^{2}} )}^{n-1}}( \frac{\sqrt{x^{2}+a^{2}}+x}{\sqrt{x^{2}+a^{2}}} ) $
$ =\frac{n{{(x+\sqrt{x^{2}+a^{2}})}^{n}}}{\sqrt{x^{2}+a^{2}}} $
$ =\frac{ny}{\sqrt{n^{2}+a^{2}}} $
BETA
