Example. Find the equations of the tangent and normal to the curve

$$
y=6 \sqrt[3]{x}-\frac{16}{3} \sqrt[4]{x}
$$

at the point with abscissa $a=1$.
Solution.

1. Find $f(1)=2 / 3$.
2. Find the derivative $f^{\prime}(1)=2 / 3$. Since $f^{\prime}(1) \neq 0$ and $f^{\prime}(1) \neq \infty$, we will use equations (1) and (2).
3. Substituting the found values $f(a)=2 / 3$ and $f^{\prime}(a)=2 / 3$ into (1) and (2), we get the equations of the tangent and normal:

$$
y=\frac{2}{3}+\frac{2}{3}(x-1), \quad y=\frac{2}{3}-\frac{3}{2}(x-1)
$$

Answer. Equation of the tangent: $2 x-3 y=0$. Equation of the normal: $9 x+6 y-13=0$.

Conditions of the Problem. Form the equations of the tangent and normal to the graph of the function $y=f(x)$ at the point with abscissa $a$.

1. $y=x-x^{2}, \quad a=1$.
2. $y=x^{2}+x+1, \quad a=-1$.
3. $y=x^{3}+x, \quad a=1$.
4. $y=\sqrt{x}-2, \quad a=4$.
5. $y=x^{2}+\sqrt{x^{3}}, \quad a=1$.
6. $y=\sqrt[3]{x^{2}}-9, \quad a=-27$.
7. $y=\frac{2+\sqrt{x}}{2-\sqrt{x}}, \quad a=9$.
8. $y=32 \sqrt[4]{x}-x, \quad a=16$.
9. $y=x^{2}-x-1, \quad a=1$.
10. $y=\frac{x^{2}-2 x+2}{x^{2}}, \quad a=2$.

Answers. 1. $x+y-1=0, x-y-1=0$. 2. $x+y=0, x-y-2=0$. 3. $4 x-y-2=0, \quad x+4 y-9=0$. 4. $x-4 y-4=0, \quad 4 x+y-16=0$. 5. $7 x-2 y-3=0,2 x+7 y-16=0$. 6. $2 x+9 y+54=0,9 x-2 y+243=0$. 7. $2 x-3 y-33=0, \quad 3 x+2 y-17=0$. 8. $y-48=0, \quad x-16=0$. 9. $x-y-2=0, \quad x+y=0 . \quad 10.2 y-1=0, \quad x-2=0$.

## 4.4. Approximate Calculations Using the Differential

Problem Statement. Calculate approximately using $\partial$ and $\emptyset$ differential the value of the function $y=f(x)$ at the point $x=a$.

Solution Plan. If the increment $\Delta x=x-a$ of the argument $x$ is small in absolute value, then

$$
f(x)=f(a+\Delta x) \approx f(a)+f^{\prime}(a) \Delta x
$$

1. Choose the point $a$ closest to $x$ and such that the values of $f(a)$ and $f^{\prime}(a)$ are easy to compute.
2. Calculate $\Delta x=x-a, f(a)$, and $f^{\prime}(a)$.
3. Using formula (1), calculate $f(x)$.