Example 2. Form the equation of the line passing through the point $B(-1,2)$ and tangent to the parabola $y=x^{2}+4 x+9$.
Solution. Point $B(-1,2)$ does not lie on the parabola, so the method described above is not applicable. Let $M_{0}\left(x_{0}, y_{0}\right)$ be the point of tangency of the sought tangent line with the parabola. Then $k_{0}=2 x_{0}+4-$ is the slope of this line. The equality $y_{0}=2+\left(2 x_{0}+4\right)\left(x_{0}+1\right)$ means that the line tangent to the parabola passes through points $B(-1,2)$ and $M_{0}\left(x_{0}, y_{0}\right)$, and the equality $y_{0}=x_{0}^{2}+4 x_{0}+9$ means that $M_{0}$ lies on the parabola. From these two equations, we find: $x_{0}^{2}+4 x_{0}+9=2+2 x_{0}^{2}+6 x_{0}+4$, i.e., $x_{0}^{2}+2 x_{0}-3=0$. Hence, $x_{0}=1$ or $x_{0}=-3 ; y_{0}=14$ or $y_{0}=6$. Thus, the points of tangency on the parabola can be $M_{0}(1,14)$ or $M_{0}^{\prime}(-3,6)$. The equations of the tangents can be formed from the two pairs of points $B, M_{0}$ and $B, M_{0}^{\prime}$. The equations themselves are $y=-2 x, y=6 x+8$.

## Exercises

Form the equations of the tangents and normals to the given curves at the given points.

1. $y=x^{3}, A(1,1)$. 2. $y=x^{2}+5, A(2,9)$. 3. $y=\sin x, A(\pi, 0)$.

Form the equations of the lines passing through the given points and tangent to the given curves.
4. $B(0,-2), y=x^{2}+x-1$.
5. $B(0,-1), y=-x^{2}-x-2$.
6. $B(1,9), y=2 x^{2}+3 x+10$.

## Answers

1. $3 x-y-2=0, x+3 y-4=0$. 2. $y=4 x+1, x+4 y-38=0$. 3. $y=\pi-x, y=x-\pi$. 4. $y=3 x-2$. 5. $y=-3 x-1$. 6. $y=7 x+2$.

## § 3. Connection between Differentiability and Continuity

$1^{\circ}$. Recall: a function that has a finite derivative at a given point is called differentiable at this point; otherwise, it is called non-differentiable.

Theorem 1. If a function is differentiable at a given point, then it is continuous at this point.

From the existence of $\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=f^{\prime}(x)$, it follows that $\lim _{\Delta x \rightarrow 0} \Delta y=0$ (continuity), because otherwise $f^{\prime}(x)=\infty$ or does not exist.

$2^{\circ}$. A function that is continuous at a given point is not necessarily differentiable at this point.

For example, the function $y=|x|$ is continuous (see Chapter V, §1), but it is not differentiable at the point $x=0$, since the algorithm with $y_{0}=y(0)=0$ leads to $y^{\prime}=|x|^{\prime}=\lim _{\Delta x \rightarrow 0} \frac{|\Delta x|}{\Delta x}$, and this limit does not exist (for $\Delta x>0$ the expression under the limit sign equals 1, and for $\Delta x<0$ it equals -1).

$3^{\circ}$. A function that is differentiable at each point of a given interval is called differentiable in this interval. A function with a continuously varying derivative (tangent) is called smooth.

The graph of the function $y=|x|$ is not smooth at $x=0$ (it has an angle).

## Exercises

Determine whether the functions $y=f(x)$ are differentiable at $x=x_{0}$.

1. $y=\sqrt[3]{x}, x_{0}=0$.
2. $y=|x-1|^{3}, x_{0}=1$.
3. $y=x^{2 / 3}, x_{0}=0$.
4. $y=x^{5 / 3}, x_{0}=0$. 5. $y=x^{3 / 5}, x_{0}=0$.

## § 4. Table of Derivatives and Rules of Differentiation

$1^{\circ}$. Differentiation of the main elementary functions is carried out using the following formulas:

$\begin{array}{ll}\text { 1) } C^{\prime}=0 . & \text { 2) } x^{\prime}=1\end{array}$
3) $\left(x^{m}\right)^{\prime}=m x^{m-1}$.
11) $(\operatorname{ctg} x)^{\prime}=-\frac{1}{\sin ^{2} x}$.
4) $\left(a^{x}\right)^{\prime}=a^{x} \ln a$.
12) $(\arcsin x)^{\prime}=\frac{1}{\sqrt{1-x^{2}}}$.
5) $\left(e^{x}\right)^{\prime}=e^{x}$.
13) $(\arccos x)^{\prime}=-\frac{1}{\sqrt{1-x^{2}}}$.
6) $\left(\log _{a} x\right)^{\prime}=\frac{1}{x \ln a}$.
14) $(\operatorname{arctg} x)^{\prime}=\frac{1}{1+x^{2}}$.
7) $(\ln x)^{\prime}=\frac{1}{x}$.
15) $(\operatorname{arcctg} x)^{\prime}=-\frac{1}{1+x^{2}}$.
8) $(\sin x)^{\prime}=\cos x$
16) $(\sqrt{x})^{\prime}=\frac{1}{2 \sqrt{x}}$.
9) $(\cos x)^{\prime}=-\sin x$.
17) $\left(\frac{1}{x}\right)^{\prime}=-\frac{1}{x^{2}}$.
10) $(\operatorname{tg} x)^{\prime}=\frac{1}{\cos ^{2} x}$.
18) $\left(\frac{1}{x^{m}}\right)^{\prime}=-\frac{m}{x^{m+1}}$.

Formulas 16)-18) are special cases of 3). All formulas must be memorized.

$2^{\circ}$. Functions composed of the main ones using arithmetic operations (addition, subtraction, multiplication, division) and composition (function of a function) are differentiated according to the rules:

1) $(C \cdot f(x))^{\prime}=C \cdot f^{\prime}(x)(C-$ constant $)$
2) $(f(x) \pm g(x))^{\prime}=f^{\prime}(x) \pm g^{\prime}(x)$
3) $(f(x) \cdot g(x))^{\prime}=f^{\prime}(x) g(x)+f(x) g^{\prime}(x)$
4) $\left(\frac{f(x)}{g(x)}\right)^{\prime}=\frac{f^{\prime}(x) g(x)-f(x) g^{\prime}(x)}{g^{2}(x)}$, in particular, $\left(\frac{1}{g(x)}\right)^{\prime}=-\frac{g^{\prime}(x)}{g^{2}(x)}$;
5) if $f(u)$ and $u=u(x)$ are differentiable functions, then the derivative of the composite function $(f[u(x)])^{\prime}=f_{u}^{\prime}[u(x)] \cdot u^{\prime}(x)$.

## Examples with Solutions

(in all examples, find $y^{\prime}$ )