Differentiation Question 463
Question: If $ y={{\tan }^{-1}}\frac{4x}{1+5x^{2}}+{{\tan }^{-1}}\frac{2+3x}{3-2x} $ , then $ \frac{dy}{dx}= $
Options:
A) $ \frac{1}{1+25x^{2}}+\frac{2}{1+x^{2}} $
B) $ \frac{5}{1+25x^{2}}+\frac{2}{1+x^{2}} $
C) $ \frac{5}{1+25x^{2}} $
D) $ \frac{1}{1+25x^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ [-3,0] $
$ ={{\tan }^{-1}}\frac{4x}{1+5x^2}+{{\tan }^{-1}}\frac{\frac{2}{3}+x}{1-\frac{2}{3}x} $
$ =6x{e^{x^{2}}}-6x{e^{x^{2}}}+\frac{1}{3}(3)-0=0$
Therefore $ \frac{dy}{dx}=\frac{5}{1+25x^{2}} $ .
BETA
