If the derivative of `tan^(-1)(a+b x)` take the value of `1` at `x=0,` prove that `1+a^2=bdot`

Explanation

`y=tan ^{-1}(a+b x) `


`Rightarrow frac{d y}{d x}=frac{1}{1+(a+b x)^{2}} frac{d}{d x}(a+b x) `


`Rightarrow frac{d y}{d x}=frac{b}{1+(a+b x)^{2}} `


`text { Atx }=0 Rightarrow frac{d y}{d x}=1 text { (given) } `


`Rightarrow 1=frac{b}{1+(a+0)^{2}}`



`therefore 1+a^{2}=b` is proved.