Question

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

Answer 1

$$\begin{gathered} \mathrm{Here},\frac{d}{dx}\left[{\tan }^{-1}\left(a+bx\right)\right]=1\mathrm{at}x=0 \\ \Rightarrow {\left[\left\{\frac{1}{1+{\left(a+bx\right)}^2}\right\}\frac{d}{dx}\left(a+bx\right)\right]}_{x=0}=1 \\ \Rightarrow {\left[\frac{1}{1+{\left(a+bx\right)}^2}\times \left(b\right)\right]}_{x=0}=1 \\ \Rightarrow \frac{b}{1+{\left(a+0\right)}^2}=1 \\ \Rightarrow b=1+a^2 \\ \therefore 1+a^2=b \end{gathered}$$