If y={\Tan}^{-1} [\frac{ \sin ^3(2 x)-3 x^2 \sin (2 x)}{3 x \sin ^2(2 x)-x^3}], then \frac{d y}{d x}= [TG EAPCET 11 May 2024 Shift 1]

Correct Answer is : 6xcos(2x)−3sin(2x)x2+sin2(2x)

\frac{6 x \cos (2 x)-3 \sin (2 x)}{x^2- \sin ^2(2 x)}

\frac{6 x \sin (2 x)-3 \cos (2 x)}{x^2+ \sin ^2(2 x)}

\frac{2 x \cos (2 x)- \sin (2 x)}{x^2+ \sin ^2(2 x)}

\frac{6 x \cos (2 x)-3 \sin (2 x)}{x^2+ \sin ^2(2 x)}

Differentiation

Section: Mathematics

Question : 3 of 15

Marks: +1, -0

If y=Tan−1[‌

sin‌3(2x)−3x2‌sin‌(2x)

3x‌sin‌2(2x)−x3

], then ‌

[TG EAPCET 11 May 2024 Shift 1]

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