608. Using the definition of the derivative, find the derivatives of the following functions:
1) $y=3 x+5$
2) $f(x)=\frac{1}{x}$ at the point $x=2$;
3) $f(x)=\sin (2 x-3)$ at the point $x=1$.
Solution. 1) Give the argument $x$ an increment $\Delta x$, then we get a new value of the function

$$
y+\Delta y=3(x+\Delta x)+5
$$

Subtract the initial value of the function 1 from this equation. We get the increment of the function $\Delta y=3 \Delta x$. Now find the limit of the ratio $\frac{\Delta y}{\Delta x}$ as $\Delta x \rightarrow 0$:

$$
\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{3 \Delta x}{\Delta x}=3
$$

i.e., $y^{\prime}=3$.

2) Calculate the increment of the function at the point $x$:

$$
\Delta y=\frac{1}{x+\Delta x}-\frac{1}{x}=-\frac{\Delta x}{x(x+\Delta x)}
$$

Find the limit of the ratio $\frac{\Delta y}{\Delta x}$ as $\Delta x \rightarrow 0$:

$$
\lim _{\Delta x \rightarrow 0} \frac{-\Delta x}{x(x+\Delta x) \cdot \Delta x}=-\lim _{\Delta x \rightarrow 0} \frac{1}{x(x+\Delta x)}=-\frac{1}{x^{2}}
$$

i.e., $f^{\prime}(x)=-\frac{1}{x^{2}}$. The derivative at the point $x=2$ is $f^{\prime}(2)=-\frac{1}{4}$.
3) $\Delta y=\sin [2(x+\Delta x)-3]-\sin (2 x-3)=2 \sin \Delta x \cos (2 x-3+\Delta x)$;

$$
\frac{\Delta y}{\Delta x}=2 \frac{\sin \Delta x}{\Delta x} \cos (2 x-3+\Delta x) \rightarrow 2 \cos (2 x-3) \text{ as } \Delta x \rightarrow 0
$$

i.e., $f^{\prime}(x)=2 \cos (2 x-3)$. The derivative at the point $x=1$ is $f^{\prime}(1)=2 \cos (-1)=2 \cos 1 \approx 2 \cos 57^{\circ} 18^{\prime} \approx 2 \cdot 0.5402=1.0804$. 609. Find the slope of the tangent line drawn to the parabola $\left.f(x)=x^{2}+4: 1\right)$ at the point $\left.(2 ; 0) ; 2\right)$ at the point $(0 ;-4)$; 3) at the point $(3 ; 5)$.

Solution. First, find the derivative at the point $x$:

$$
\begin{gathered}
\Delta y=(x+\Delta x)^{2}+4-\left(x^{2}+4\right)=2 x \Delta x+\Delta x^{2} \\
\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{(2 x+\Delta x) \cdot \Delta x}{\Delta x}=\lim _{\Delta x \rightarrow 0}(2 x+\Delta x)=2 x
\end{gathered}
$$

i.e., $f^{\prime}(x)=2 x$.

The slopes of the tangent lines to the given curve at the specified points are equal to the values of the derivative respectively at $x=2, x=0$, and $x=3$, i.e.,

$$
f^{\prime}(2)=2 \cdot 2=4 ; f^{\prime}(0)=2 \cdot 0=0 ; f^{\prime}(3)=2 \cdot 3=6
$$