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Problem Given y = arccos ( 1 − 3 x 1 + 3 x ) y=\arccos\!\left(\dfrac{1-3x}{1+3x}\right) y = arccos ( 1 + 3 x 1 − 3 x ) , which of the following is d y d x \dfrac{dy}{dx} d x d y ?
a) − 6 ( 1 + 3 x ) 2 1 − ( 1 − 3 x 1 + 3 x ) 2 -\dfrac{6}{(1+3x)^{2}\,\sqrt{1-\left(\dfrac{1-3x}{1+3x}\right)^{2}}} − ( 1 + 3 x ) 2 1 − ( 1 + 3 x 1 − 3 x ) 2 6
b) 6 ( 1 + 3 x ) 2 1 − ( 1 − 3 x 1 + 3 x ) 2 \dfrac{6}{(1+3x)^{2}\,\sqrt{1-\left(\dfrac{1-3x}{1+3x}\right)^{2}}} ( 1 + 3 x ) 2 1 − ( 1 + 3 x 1 − 3 x ) 2 6
c) 6 ( 1 + 3 x ) 1 − ( 1 − 3 x 1 + 3 x ) 2 \dfrac{6}{(1+3x)\,\sqrt{1-\left(\dfrac{1-3x}{1+3x}\right)^{2}}} ( 1 + 3 x ) 1 − ( 1 + 3 x 1 − 3 x ) 2 6
d) − 1 1 − ( 1 − 3 x 1 + 3 x ) 2 -\dfrac{1}{\sqrt{1-\left(\dfrac{1-3x}{1+3x}\right)^{2}}} − 1 − ( 1 + 3 x 1 − 3 x ) 2 1
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