383. At what angle do the curves $y=2^{x}$ and $y=$ $=\sqrt{x+1}$ intersect?
Solution. Let's find the points of intersection of the given curves; by solving the equation $2^{x}=\sqrt{x+1}$, we get $x_{1}=-0.5, x_{2}=0$.

The angle between the curves is understood as the angle between the tangents at the point of intersection of the curves. The magnitude of the angle is found using the known formula $\operatorname{tg} \varphi=\frac{k_{2}-k_{1}}{1+k_{1} k_{2}}$, where $k_{1}, k_{2}$ are the slopes of the tangents at the point of their intersection.

Let's find $y_{1}^{\prime}=2^{x} \ln 2, y_{2}^{\prime}=\frac{1}{2 \sqrt{x+1}}$. We will calculate the values of the slopes at the points $x=-0.5$ and $x=0$:

$$
\begin{gathered}
k_{1}(-0.5)=2^{-0.5} \ln 2=\frac{\ln 2}{\sqrt{2}}, k_{2}(-0.5)=\frac{1}{2 \sqrt{0.5}}=\frac{\sqrt{2}}{2} ; k_{1}(0)=\ln 2 \\
k_{2}(0)=\frac{1}{2}
\end{gathered}
$$

Thus, at the point $x=-0.5$ we have

$$
\operatorname{tg} \varphi=\frac{\frac{\sqrt{2}}{2}-\frac{\ln 2}{\sqrt{2}}}{1+\frac{\sqrt{2}}{2} \cdot \frac{\ln 2}{\sqrt{2}}}=\frac{(1-\ln 2) \sqrt{2}}{2+\ln 2}, \text { hence } \varphi=\operatorname{arctg} \frac{(1-\ln 2) \sqrt{2}}{2+\ln 2}
$$

At the point $x=0$ we have

$$
\operatorname{tg} \varphi=\frac{\frac{1}{2}-\ln 2}{1+\frac{1}{2} \ln 2}=\frac{1-2 \ln 2}{2+\ln 2}, \text { i.e. } \varphi=\operatorname{arctg} \frac{1-2 \ln 2}{2+\ln 2}
$$