TS EAMCET · Maths · Differentiation
If \(y=\tan ^{-1}\left[\frac{\sin ^3(2 x)-3 x^2 \sin (2 x)}{3 x \sin ^2(2 x)-x^3}\right]\), then \(\frac{d y}{d x}=\)
- A \(\frac{6 x \cos (2 x) 3 \sin (2 x)}{x^2 \sin ^2(2 x)}\)
- B \(\frac{6 x \sin (2 x)-3 \cos (2 x)}{x^2+\sin ^2(2 x)}\)
- C \(\frac{2 x \cos (2 x)-\sin (2 x)}{x^2+\sin ^2(2 x)}\)
- D \(\frac{6 x \cos (2 x)-3 \sin (2 x)}{x^2+\sin ^2(2 x)}\)
Answer & Solution
Correct Answer
(D) \(\frac{6 x \cos (2 x)-3 \sin (2 x)}{x^2+\sin ^2(2 x)}\)
Step-by-step Solution
Detailed explanation
\(y=\tan ^{-1}\left[\frac{\sin ^3(2 x)-3 x^2 \sin 2 x}{3 x \sin ^2(2 x)-x^3}\right]\) Dividing numerator and denominator by \(x^3\)…
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