Differentiation Question 409
Question: $ \frac{d}{dx}( \frac{{{\cot }^{2}}x-1}{{{\cot }^{2}}x+1} )= $
Options:
A) $ -\sin 2x $
B) $ 2\sin 2x $
C) $ 2\cos 2x $
D) $ -2\sin 2x $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{d}{dx}[ \frac{{{\cot }^{2}}x-1}{{{\cot }^{2}}x+1} ]=\frac{d}{dx}[ \frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x+{{\sin }^{2}}x} ] $
$ =\frac{d}{dx}[\cos 2x]=-2\sin 2x $ .
BETA
