Example 5 (to $5^{\circ}$ ). Find $y^{\prime \prime}$, if $x=\ln t, y=\sin 2 t$.
Solution. To determine $y^{\prime \prime}$, we will use the second formula. First, we find: $x_{t}^{\prime}=\frac{1}{t}, x_{t^{2}}^{\prime \prime}=-\frac{1}{t^{2}}, y_{t}^{\prime}=2 \cos 2 t, y_{t^{2}}^{\prime \prime}=-4 \sin 2 t$.

Therefore,

$$
y_{x^{2}}^{\prime \prime}=\frac{-4 \sin 2 t \cdot \frac{1}{t}+\frac{1}{t^{2}} \cdot 2 \cos 2 t}{\left(\frac{1}{t}\right)^{3}}=-4 t^{2} \cdot \sin 2 t+2 t \cos 2 t
$$

## Exercises

Find the derivatives $y^{\prime}=\frac{d y}{d x}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-083.jpg?height=95&width=892&top_left_y=674&top_left_x=129)
4. $x=e^{-t}, y=e^{2 t}$
5. $x=\sqrt{t^{2}+1}, y=\frac{t-1}{\sqrt{t^{2}+1}}$
6. $x=\arccos \frac{1}{\sqrt{1+t^{2}}}, y=\arcsin \frac{t}{\sqrt{1+t^{2}}}$. 7. $y=(\operatorname{arctg} x)^{x}$.
8. $y=x^{x^{2}}$.

Find the derivatives of functions given implicitly by equations.
9. $x^{3} y^{3}+5 x y+4=0$. 10. $x \sin y+y \sin x=0$.
11. $x^{y}-y^{x}=0$. 12. $x=e^{-t} \sin t, \quad y=e^{t} \cos t$.
13. $x=\ln t, \quad y=\sin 2 t$.

Find derivatives of any order.
14. $y=e^{a x}$. 15. $y=\sin a x+\cos b x$.
16. $y=x e^{x}$.
17. $y=\frac{1}{a x+b}$
18. $y=\frac{x}{x^{2}-1}$

## Answers

1. $y^{\prime}=\frac{x^{2}-a y}{a x-y^{2}}$. 2. $y^{\prime}=\frac{y\left(1-x^{2}-y^{2}\right)}{x\left(1+x^{2}+y^{2}\right)}$. 3. $y^{\prime}=\frac{1}{x+y-1}=\frac{1}{e^{y}-1}$.
2. $y^{\prime}=-2 e^{3 t}$. 5. $y^{\prime}=\frac{t+1}{t\left(t^{2}+1\right)}$. 6. $y^{\prime}=-1$ for $t0$. 7. $y^{\prime}=(\operatorname{arctg} x)^{x}\left(\ln \operatorname{arctg} x+\frac{x}{\left(1+x^{2}\right) \operatorname{arctg} x}\right)$. 8. $y^{\prime}=$ $=x^{x^{2}+1}(1+2 \ln x) . \mathbf{9} . y^{\prime}=-\frac{3 x^{2} y^{2}+5 y}{2 x^{3} y+5 x} . \mathbf{1 0} \cdot y^{\prime}=\frac{-(y \cos x+\sin y)}{x \cos y+\sin x}$. 11. $y^{\prime}=\frac{y(y-x \ln y)}{x(x-y \ln x)}$. 12. $y^{\prime}=e^{2 t}$. 13. $y^{\prime}=2 t \cos 2 t$. 14. $y^{(n)}=a^{n} e^{a x}$. 15. $y^{(n)}=a^{n} \sin \left(a x+\frac{n \pi}{2}\right)+b^{n} \cos \left(b x+\frac{n \pi}{2}\right)$. 16. $y^{(n)}=(x+n) e^{x}$. 17. $y^{(n)}=\frac{(-1)^{n} a^{n} n!}{(a x+b)^{n+1}} \cdot$ 18. $y^{(n)}=(-1)^{n} \frac{n!}{2}\left(\frac{1}{(x+1)^{n+1}}+\frac{1}{(x-1)^{n+1}}\right)$.

## § 8. Main Theorems of Differential Calculus

$1^{\circ}$. A point $x_{0}$ is called a point of relative maximum (minimum) of the function $f(x)$ if in some neighborhood of this point the inequality $f(x)f\left(x_{0}\right)\right)$ holds for $x \neq x_{0}$. Points of relative maximum and minimum of $f(x)$ are called points of extremum, and the values at these points are called extremal values.

Figure 7.2 shows the graph of a function with three extrema.

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-083.jpg?height=210&width=456&top_left_y=597&top_left_x=1408)

Fig. 7.2

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-083.jpg?height=183&width=424&top_left_y=624&top_left_x=2005)

Fig. 7.3
Theorem 2 (Fermat). If a differentiable function $f(x)$ has an extremum at a point $x_{0}$, then $f^{\prime}\left(x_{0}\right)=0$.

At the point of extremum, the tangent is horizontal, and for it $k=$ $=\operatorname{tg} \alpha=f^{\prime}\left(x_{0}\right)=0$.

Theorem 3 (Rolle). Suppose that the function $f(x)$ is continuous on the interval $[a ; b]$, differentiable in the interval $(a ; b)$, and $f(a)=f(b)$. Then there exists at least one point $c \in(a ; b)$ such that $f^{\prime}(\mathrm{c})=0$.

Such a function is either constant, i.e., $f(x)=C$, and then $f^{\prime}(x) \equiv 0$, $x \in[a ; b]$, or it has at least one extremum point $c \in(a ; b)$, and then $f^{\prime}(c)=0$.

Theorem 4 (Lagrange). Suppose that the function $f(x)$ is continuous on the interval $[a ; b]$ and differentiable in the interval $(a ; b)$. Then there exists at least one point $c \in(a ; b)$ such that $f(b)-f(a)=f^{\prime}(c)(b-a)$, or $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$.

Geometrically, Theorem 4 means that on the curve there exists at least one point where the tangent is parallel to the chord $A B, f^{\prime}(c)-$ its slope (Fig. 7.3).

Theorem 5 (Cauchy). Let $f(x)$ and $g(x)$ be two functions continuous on the interval $[a ; b]$ and differentiable in the interval $(a ; b)$, with $g^{\prime}(x) \neq 0, x \in(a ; b)$. Then there exists at least one point such that

$$
\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f^{\prime}(c)}{g^{\prime}(c)}
$$

## Examples with Solutions