Differentiation Question 350
Question: $ \frac{d}{dx}{{\tan }^{-1}}( \frac{ax-b}{bx+a} )= $
Options:
A) $ \frac{1}{1+x^{2}}-\frac{a^{2}}{a^{2}+b^{2}} $
B) $ \frac{-1}{1+x^{2}}-\frac{a^{2}}{a^{2}+b^{2}} $
C) $ \frac{1}{1+x^{2}}+\frac{a^{2}}{a^{2}+b^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{d}{dx}{{\tan }^{-1}}( \frac{ax-b}{bx+a} )=\frac{1}{1+{{( \frac{ax-b}{bx+a} )}^{2}}}. $
$ \frac{d}{dx}( \frac{ax-b}{bx+a} ) $
$ =\frac{a^{2}+b^{2}}{a^{2}+b^{2}+a^{2}x^{2}+b^{2}x^{2}}=\frac{1}{1+x^{2}} $ .
BETA
