The derivative of \(e^{a x} \cos b x\) with respect \(x\) is \(r^{a x} \cos b x \tan ^{-1} \frac{b}{a}\). When \(a>0, b>0\), then value of \(r\), is

Given, \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\mathrm{ax}} \cos \mathrm{bx}\right)=\mathrm{re}^{\mathrm{ax}} \cos (\mathrm{bx}+\alpha)\), then \(\mathrm{r}=\) ?
Let \(\quad y=e^{a x} \cdot \cos b x\)
\(\frac{d y}{d x}=a e^{a x} \cdot \cos b x-b e^{a x} \cdot \sin b x\)
\(\frac{d y}{d x}=e^{a x}(a \cos b x-b \sin b x)\)
Let \(\left.\quad \begin{array}{l}\mathrm{a}=\mathrm{r} \cos \alpha \\ \mathrm{b}=\mathrm{r} \sin \alpha\end{array}\right\} \quad \text{...(i)}\)
Then, \(\frac{d y}{d x}=e^{a x} \cdot r\{\cos b x \cdot \cos \alpha-\sin b x \cdot \sin \alpha\}\)
\[
\frac{d y}{d x}=e^{a x} \cdot r \cos (b x+\alpha) \quad \text{...(ii)}
\]
Where, \(\tan \alpha=\frac{\mathrm{b}}{\mathrm{a}} \Rightarrow \alpha=\tan ^{-1}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\) and
\[
\begin{aligned}
\mathrm{r}^{2}\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right) &=\mathrm{a}^{2}+\mathrm{b}^{2} \\
\mathrm{r}^{2} &=\mathrm{a}^{2}+\mathrm{b}^{2} \\
\mathrm{r} &=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}
\end{aligned}
\]