Question

Y=tan‐¹ (x/1+6x²) then dy/dx

Answer 1

Step-by-step explanation:

We have,

$y = \tan^{ - 1} \bigg( \frac{x}{1 + 6 {x}^{2} } \bigg)$

$\implies \: y = \tan^{ - 1} \bigg( \frac{3x - 2x}{1 + 3x.2x } \bigg)$

$\implies \: y = \tan^{ - 1} (3x) - \tan ^{ - 1} (2x)$

$\implies \: \frac{dy}{dx} = \frac{3}{1 + 9 {x}^{2} } - \frac{2}{1 + 4 {x}^{2} }$