Question:
If the derivative of $\tan ^{-1}(a+b x)$ takes the value 1 at $x=0$, prove that $1+a^{2}=b$.
Solution:
$y=\tan ^{-1}(a+b x)$
And $y^{\prime}(0)=1$
Now
$\frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1}(a+b x)\right)$
$\frac{d y}{d x}=\frac{b}{1+(a+b x)^{2}}$
At $x=0$
$\frac{d y}{d x}=\frac{b}{1+(a+b(0))^{2}}$
$\frac{b}{1+a^{2}}=1$
$\Rightarrow b=1+a^{2}$
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