AP EAMCET · Maths · Inverse Trigonometric Functions
\(\text { If } y=\operatorname{Tan}^{-1}\left(\frac{x}{1+2 x^2}\right)+\operatorname{Tan}^{-1}\left(\frac{x}{1+6 x^2}\right) \text {, then } \frac{d y}{d x}=\)
- A \(\frac{4}{16 x^2+1}-\frac{3}{9 x^2+1}\)
- B \(\frac{3}{9 x^2+1}-\frac{1}{x^2+1}\)
- C \(\frac{3}{9 x^2+1}-\frac{2}{4 x^2+1}\)
- D \(\frac{1}{9 x^2+1}-\frac{1}{x^2+1}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{9 x^2+1}-\frac{1}{x^2+1}\)
Step-by-step Solution
Detailed explanation
\(y=\operatorname{Tan}^{-1}(2x)-\operatorname{Tan}^{-1}(x)+\operatorname{Tan}^{-1}(3x)-\operatorname{Tan}^{-1}(2x)\) \(y=\operatorname{Tan}^{-1}(3x)-\operatorname{Tan}^{-1}(x)\) \(\frac{dy}{dx}=\frac{d}{dx}(\operatorname{Tan}^{-1}(3x))-\frac{d}{dx}(\operatorname{Tan}^{-1}(x))\)…
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