Example. Find the $n$-th order derivative of the function $y=3^{2 x+5}$.






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Solution.

1. Let's find sequentially

$$
\begin{aligned}
& y^{\prime}(x)=\left(3^{2 x+1}\right)^{\prime}=3^{2 x+1}(\ln 3) 2 \\
& y^{\prime \prime}(x)=y^{\prime}(x)^{\prime}=\left(3^{2 x+1}(\ln 3) 2\right)^{\prime}=3^{2 x+1}(\ln 3)^{2} 2^{2} \\
& y^{\prime \prime \prime}(x)=y^{\prime \prime}(x)^{\prime}=\left(3^{2 x+1}(\ln 3)^{2} 2^{2}\right)^{\prime}=3^{2 x+1}(\ln 3)^{3} 2^{3}
\end{aligned}
$$

Analyzing these expressions, we hypothesize that

$$
y^{(n)}(x)=3^{2 x+1}(\ln 3)^{n} 2^{n}
$$

2. We will prove this formula by mathematical induction. First, check that it holds for $n=1$; i.e.,

$$
y^{(1)}(x)=3^{2 x+1}(\ln 3) 2=y^{\prime}(x)
$$

Differentiating $f^{(n)}$ is equivalent to replacing $n$ with $n+1$, i.e.,

$$
y^{(n)}(x)^{\prime}=\left(3^{2 x+1}(\ln 3)^{n} 2^{n}\right)^{\prime}=3^{2 x+1}(\ln 3)^{n+1} 2^{n+1}=y^{(n+1)}(x)
$$

Answer. $y^{(n)}(x)=3^{2 x+1}(2 \ln 3)^{n}$.

Conditions of the Problems. Find the $n$-th order derivatives of the given functions.

1. $y=\sin 2 x+\cos 3 x$. 2. $y=\sin (3 x+1)+\cos 2 x$. 3. $y=2^{3 x}$.
2. $y=\ln (2 x+4)$.
3. $y=\frac{x}{x+1}$.
4. $y=\frac{x+1}{2 x+3}$.
5. $y=3^{2 x+1}$.
6. $y=\ln (3 x+1)$.
7. $y=5^{2 x+4}$.
8. $y=\sqrt{x}$.

## Answers.

1. $y^{(n)}=2^{n} \sin \left(2 x+\frac{n \pi}{2}\right)+3^{n} \cos \left(3 x+\frac{n \pi}{2}\right)$.
2. $y^{(n)}=3^{n} \sin \left(3 x+1+\frac{n \pi}{2}\right)+2^{n} \cos \left(2 x+\frac{n \pi}{2}\right)$.
3. $y^{(n)}=2^{3 x}(3 \ln 2)^{n} . \quad$ 4. $y^{(n)}=\frac{(-1)^{n-1} 2^{n}(n-1)!}{(2 x+4)^{n}}$.
4. $y^{(n)}=\frac{(-1)^{n-1} n!}{(x+1)^{n+1}}$.
5. $y^{(n)}=\frac{(-1)^{n-1} 2^{n-1} n!}{(2 x+3)^{n+1}}$.
6. $y^{(n)}=3^{2 x+1}(2 \ln 3)^{n}$.
7. $y^{(n)}=\frac{(-1)^{n-1} 3^{n}(n-1)!}{(3 x+1)^{n}}$.
8. $y^{(n)}=5^{2 x+4}(2 \ln 5)^{n}$.
9. $y^{(n)}=\frac{(-1)^{n-1}\left(2^{n}-3\right)!!}{2^{n}(\sqrt{x})^{2 n-1}}$.

## 4.9. Leibniz's Formula

Problem Statement. Find the $n$-th order derivative of the function $y=u(x) v(x)$.

Plan of Solution. If the functions $u(x)$ and $v(x)$ have derivatives up to the $n$-th order inclusive, then the following formulas are valid:

$$
\begin{aligned}
& (u v)^{\prime \prime}=u^{\prime \prime} v+2 u^{\prime} v^{\prime}+u v^{\prime \prime} \\
& (u v)^{\prime \prime \prime}=u^{\prime \prime \prime} v+3 u^{\prime \prime} v^{\prime}+3 u^{\prime} v^{\prime \prime}+u v^{\prime \prime \prime} \\
& (u v)^{(4)}=u^{(4)} v+4 u^{\prime \prime \prime} v^{\prime}+6 u^{\prime \prime} v^{\prime \prime}+4 u^{\prime} v^{\prime \prime \prime}+u v^{(4)}
\end{aligned}
$$

$$
\begin{aligned}
(u v)^{(n)}=u^{(n)} v+n u^{(n-1)} v^{\prime}+\frac{n(n-1)}{1 \cdot 2} & u^{(n-2)} v^{\prime \prime}+\ldots+u v^{(n)}= \\
= & \sum_{k=0}^{n} C_{n}^{k} u^{(n-k)} v^{(k)}
\end{aligned}
$$

where $u^{(0)}=u, v^{(0)}=v$ and $C_{n}^{k}=\frac{n!}{k!(n-k)!}$ are binomial coefficients.

The formula (1) for the $n$-th derivative of a product is called Leibniz's formula.

Consequently, to determine the $n$-th order derivative of a function of the form $y=u(x) v(x)$, we need to compute all derivatives (up to the $n$-th order inclusive) of each of the functions $u(x)$ and $v(x)$, the binomial coefficients $C_{n}^{k}$, and use Leibniz's formula.