If y={\Tan}^{-1} (\frac{x}{1+2 x^2}) +{\Tan}^{-1} (\frac{x}{1+6 x^2}), then \frac{d y}{d x}= [AP EAPCET 22 May 2025 S2]

Correct Answer is : 39x2+1−1x2+1

\frac{4}{16 x^2+1}-\frac{3}{9 x^2+1}

\frac{3}{9 x^2+1}-\frac{1}{x^2+1}

\frac{3}{9 x^2+1}-\frac{2}{4 x^2+1}

\frac{1}{9 x^2+1}-\frac{1}{x^2+1}

Inverse Trigonometric Functions

Section: Mathematics

Question : 10 of 33

Marks: +1, -0

If y=Tan−1(‌

1+2x2

)+Tan−1(‌

1+6x2

), then ‌

[AP EAPCET 22 May 2025 S2]

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