Example 5. Find the tenth derivative for the function $y=$ $=e^{x}\left(x^{3}-2\right)$.
Solution. Applying the Leibniz formula, we get

$$
\begin{aligned}
y^{(10)}= & {\left[e^{x}\left(x^{3}-2\right)\right]^{(10)}=\left(e^{x}\right)^{(10)}\left(x^{3}-2\right)+10\left(e^{x}\right)^{(9)}\left(x^{3}-2\right)^{\prime}+} \\
& +\frac{10 \cdot 9}{2}\left(e^{x}\right)^{(8)}\left(x^{3}-2\right)^{\prime \prime}+\frac{10 \cdot 9 \cdot 8}{2 \cdot 3}\left(e^{x}\right)^{(7)}\left(x^{3}-2\right)^{\prime \prime \prime}
\end{aligned}
$$

All subsequent terms are zero because all higher-order derivatives of the function $x^{3}-2$, starting from the fourth, are zero. Since the derivative of any order of $e^{x}$ is $e^{x}$, we have

$$
\begin{gathered}
y^{(10)}=e^{x}\left(x^{3}-2\right)+30 e^{x} x^{2}+45 e^{x} \cdot 6 x+120 e^{x} \cdot 6 \\
y^{(10)}=e^{x}\left(x^{3}+30 x^{2}+270 x+718\right)
\end{gathered}
$$

## Problems

Find the second-order derivatives of the functions:

1. $y=\left(x^{2}-1\right)^{2}$.
2. $y=\left(a x^{2}-b\right)^{2}$.
3. $y=a^{x}$.
4. $y=\ln x$.
5. $y=\tan x$.
6. $y=\cot x$.

Find the third-order derivatives of the functions:
7. $r=a(\varphi-\sin \varphi)$.
8. $r=a(1-\cos \varphi)$.
9. $s=a \sin 4 t$.
10. $s=a \cos 3 t$.

## Answers

1. $4\left(3 x^{2}-1\right)$.
2. $4 a\left(3 a x^{2}-b\right)$.
3. $a^{x} \ln ^{2} a$.
4. $-\frac{1}{x^{2}}$.
5. $\frac{2 \sin x}{\cos ^{3} x}$.
6. $\frac{2 \cos x}{\sin ^{3} x}$. 7. $a \cos \varphi .8 .-a \sin \varphi .9 .-64 a \cos 4 t$. 10. $27 a \sin 3 t$.

## § 2.7. Derivatives of Hyperbolic Functions and Functions Given Parametrically

The expressions $\frac{e^{x}-e^{-x}}{2}, \frac{e^{x}+e^{-x}}{2}$ and their ratios are called, respectively, the hyperbolic sine, cosine, tangent, and cotangent, and are denoted as:

$$
\begin{aligned}
\operatorname{sh} x=\frac{e^{x}-e^{-x}}{2} ; \operatorname{ch} x & =\frac{e^{x}+e^{-x}}{2} ; \operatorname{th} x=\frac{\operatorname{sh} x}{\operatorname{ch} x}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \\
\operatorname{cth} x & =\frac{\operatorname{ch} x}{\operatorname{sh} x}=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}
\end{aligned}
$$

Basic formulas for hyperbolic functions:

$$
\begin{gathered}
\operatorname{ch}^{2} x-\operatorname{sh}^{2} x=1 \\
\operatorname{ch}^{2} x+\operatorname{sh}^{2} x=\operatorname{ch} 2 x \\
\operatorname{sh} 2 x=2 \operatorname{sh} x \operatorname{ch} x
\end{gathered}
$$

1. Derivatives of hyperbolic functions:

$$
\begin{gathered}
(\operatorname{sh} x)^{\prime}=\operatorname{ch} x ;(\operatorname{ch} x)^{\prime}=\operatorname{sh} x \\
(\operatorname{th} x)^{\prime}=\frac{1}{\operatorname{ch}^{2} x} ;(\operatorname{cth} x)^{\prime}=-\frac{1}{\operatorname{sh}^{2} x}
\end{gathered}
$$

2. The derivative $y_{x}^{\prime}$ of a function given parametrically

$$
x=\varphi_{1}(t), y=\varphi_{2}(t)
$$

is found using the formula

$$
y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{i}^{\prime}}=\frac{\varphi_{2}^{\prime}}{\varphi_{1}^{\prime}} .
$$

3. The derivative of the function $y=u^{v}$, where $u$ and $v$ are differentiable functions of $x$, is found using preliminary logarithmic differentiation.

Example 1. Find the derivative of the function $y=\operatorname{sh}^{2} x$. $\mathbf{P e \boldsymbol { u }} \mathbf{4} \mathbf{e}$. Differentiating the composite function and using the first
of formulas (2.21), we get

$$
y^{\prime}=2 \operatorname{sh} x \cdot(\operatorname{sh} x)^{\prime}=2 \operatorname{sh} x \operatorname{ch} x=\operatorname{sh} 2 x
$$