Question

Find d/dx tan⁻¹ a+bx/b-ax

Answer 1

we have to find the value of $\frac{d}{dx}tan^{-1}\frac{a+bx}{b-ax}$

d/dx [tan^-1{(a + bx)/(b - ax)}] = 1/[1 + (a + bx)²/(b - ax)²] × d{(a + bx)/(b - ax)}/dx

= (b - ax)²/[(b - ax)² + (a + bx)²] × [(b - ax) × b - (a + bx) × (-a)]/(b - ax)²

= (b - ax)²/[b² + a²x² + a² + b²x²] × [{b² + a²}/(b - ax)²}]

= (b - ax)²/[(x² + 1)(a² + b²)] × [(b² + a²)/(b - ax)²]

= (b - ax)²/(x² + 1) × 1/(b - ax)²

= 1/(1 + x²)

hence answer is 1/(1 + x²)