Question

If the derivative of tan⁻¹ (a + bx) takes the value 1 at x = 0, prove that 1 + a² = b ?

Answer 1

\[\text{Here}, \frac{d}{dx}\left[ \tan^{- 1} \left( a + bx \right) \right] = 1 \text{ at }x = 0\]

\[ \Rightarrow \left[ \left\{ \frac{1}{1 + \left( a + bx \right)^2} \right\}\frac{d}{dx}\left( a + bx \right) \right]_{x = 0} = 1\]

\[ \Rightarrow \left[ \frac{1}{1 + \left( a + bx \right)^2} \times \left( b \right) \right]_{x = 0} = 1\]

\[ \Rightarrow \frac{b}{1 + \left( a + 0 \right)^2} = 1\]

\[ \Rightarrow b = 1 + a^2 \]

\[ \therefore 1 + a^2 = b\]