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1. $18.01$ Single Variable Calculus
Fall 2006
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Lecture 3 Derivatives of Products, Quotients, Sine, and Cosine
There are two kinds of derivative formulas:
- Specific Examples: $\frac{d}{d x} x^{n}$ or $\frac{d}{d x}\left(\frac{1}{x}\right)$
- General Examples: $(u+v)^{\prime}=u^{\prime}+v^{\prime}$ and $(c u)=c u^{\prime}$ (where $c$ is a constant)
A notational convention we will use today is:
$$
(u+v)(x)=u(x)+v(x) ; \quad u v(x)=u(x) v(x)
$$
Proof of $(u+v)=u^{\prime}+v^{\prime}$. (General)
Start by using the definition of the derivative.
$$
\begin{aligned}
(u+v)^{\prime}(x) &=\lim {\Delta x \rightarrow 0} \frac{(u+v)(x+\Delta x)-(u+v)(x)}{\Delta x} \ &=\lim {\Delta x \rightarrow 0} \frac{u(x+\Delta x)+v(x+\Delta x)-u(x)-v(x)}{\Delta x} \
&=\lim _{\Delta x \rightarrow 0}\left{\frac{u(x+\Delta x)-u(x)}{\Delta x}+\frac{v(x+\Delta x)-v(x)}{\Delta x}\right} \
(u+v)^{\prime}(x) &=u^{\prime}(x)+v^{\prime}(x)
\end{aligned}
$$
Follow the same procedure to prove that $(c u)^{\prime}=c u^{\prime}$.
Derivatives of $\sin x$ and $\cos x$. (Specific)
Last time, we computed
$$
\begin{aligned}
\lim {x \rightarrow 0} \frac{\sin x}{x} &=1 \ \left.\frac{d}{d x}(\sin x)\right|{x=0} &=\lim {\Delta x \rightarrow 0} \frac{\sin (0+\Delta x)-\sin (0)}{\Delta x}=\lim {\Delta x \rightarrow 0} \frac{\sin (\Delta x)}{\Delta x}=1 \
\left.\frac{d}{d x}(\cos x)\right|{x=0} &=\lim {\Delta x \rightarrow 0} \frac{\cos (0+\Delta x)-\cos (0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\cos (\Delta x)-1}{\Delta x}=0
\end{aligned}
$$
So, we know the value of $\frac{d}{d x} \sin x$ and of $\frac{d}{d x} \cos x$ at $x=0$. Let us find these for arbitrary $x$.
$$
\frac{d}{d x} \sin x=\lim _{\Delta x \rightarrow 0} \frac{\sin (x+\Delta x)-\sin (x)}{\Delta x}
$$
Recall:
$$
\sin (a+b)=\sin (a) \cos (b)+\sin (b) \cos (a)
$$
So,
$$
\begin{aligned}
\frac{d}{d x} \sin x &=\lim {\Delta x \rightarrow 0} \frac{\sin x \cos \Delta x+\cos x \sin \Delta x-\sin (x)}{\Delta x} \ &=\lim {\Delta x \rightarrow 0}\left[\frac{\sin x(\cos \Delta x-1)}{\Delta x}+\frac{\cos x \sin \Delta x}{\Delta x}\right] \
&=\lim {\Delta x \rightarrow 0} \sin x\left(\frac{\cos \Delta x-1}{\Delta x}\right)+\lim {\Delta x \rightarrow 0} \cos x\left(\frac{\sin \Delta x}{\Delta x}\right)
\end{aligned}
$$
Since $\frac{\cos \Delta x-1}{\Delta x} \rightarrow 0$ and that $\frac{\sin \Delta x}{\Delta x} \rightarrow 1$, the equation above simplifies to
$$
\frac{d}{d x} \sin x=\cos x
$$
A similar calculation gives
$$
\frac{d}{d x} \cos x=-\sin x
$$
$$
(u v)^{\prime}=u^{\prime} v+u v^{\prime}
$$
Proof:
$$
(u v)^{\prime}=\lim {\Delta x \rightarrow 0} \frac{(u v)(x+\Delta x)-(u v)(x)}{\Delta x}=\lim {\Delta x \rightarrow 0} \frac{u(x+\Delta x) v(x+\Delta x)-u(x) v(x)}{\Delta x}
$$
Now obviously,
$$
u(x+\Delta x) v(x)-u(x+\Delta x) v(x)=0
$$
so adding that to the numerator won’t change anything.
$$
(u v)^{\prime}=\lim _{\Delta x \rightarrow 0} \frac{u(x+\Delta x) v(x)-u(x) v(x)+u(x+\Delta x) v(x+\Delta x)-u(x+\Delta x) v(x)}{\Delta x}
$$
We can re-arrange that expression to get
$$
(u v)^{\prime}=\lim _{\Delta x \rightarrow 0}\left(\frac{u(x+\Delta x)-u(x)}{\Delta x}\right) v(x)+u(x+\Delta x)\left(\frac{v(x+\Delta x)-v(x)}{\Delta x}\right)
$$
Remember, the limit of a sum is the sum of the limits.
$$
\begin{gathered}
{\left[\lim {\Delta x \rightarrow 0} \frac{u(x+\Delta x)-u(x)}{\Delta x}\right] v(x)+\lim {\Delta x \rightarrow 0}\left(u(x+\Delta x)\left[\frac{v(x+\Delta x)-v(x)}{\Delta x}\right]\right)} \
(u v)^{\prime}=u^{\prime}(x) v(x)+u(x) v^{\prime}(x)
\end{gathered}
$$
Note: we also used the fact that
$$
\lim _{\Delta x \rightarrow 0} u(x+\Delta x)=u(x) \quad \text { (true because } u \text { is continuous) }
$$
This proof of the product rule assumes that $u$ and $v$ have derivatives, which implies both functions are continuous.
- A sequence of numbers $x_{1}, x_{2}, \ldots, x_{n}, \ldots$ satisfies the recurrence r tion $x_{n+1}=a x_{n}+b x_{n-1}$ for $n \geq 1$, where $a$ and $b$ are constants. Pruve that
$$
\left[\begin{array}{c}
x_{n+1} \
x_{n}
\end{array}\right]=A\left[\begin{array}{c}
x_{n} \
x_{n-1}
\end{array}\right]
$$
35
2.4. PROBLEMS
where $A=\left[\begin{array}{ll}a & b \ 1 & 0\end{array}\right]$ and hence express $\left[\begin{array}{c}x_{n+1} \ x_{n}\end{array}\right]$ in terms of $\left[\begin{array}{l}x_{1} \ x_{0}\end{array}\right]$. If $a=4$ and $b=-3$, use the previous question to find a formula for $x_{n}$ in terms of $x_{1}$ and $x_{0}$.
[Answer:
$$
\left.x_{n}=\frac{3^{n}-1}{2} x_{1}+\frac{3-3^{n}}{2} x_{0}\right]
$$ - Let $A=\left[\begin{array}{cc}2 a & -a^{2} \ 1 & 0\end{array}\right]$.
(a) Prove that
$$
A^{n}=\left[\begin{array}{cc}
(n+1) a^{n} & -n a^{n+1} \
n a^{n-1} & (1-n) a^{n}
\end{array}\right] \quad \text { if } n \geq 1
$$
(b) A sequence $x_{0}, x_{1}, \ldots, x_{n}, \ldots$ satisfies the recurrence relation $x_{n+1}=$ $2 a x_{n}-a^{2} x_{n-1}$ for $n \geq 1$. Use part (a) and the previous question to prove that $x_{n}=n a^{n-1} x_{1}+(1-n) a^{n} x_{0}$ for $n \geq 1$. - Let $A=\left[\begin{array}{ll}a & b \ c & d\end{array}\right]$ and suppose that $\lambda_{1}$ and $\lambda_{2}$ are the roots of the quadratic polynomial $x^{2}-(a+d) x+a d-b c$. ( $\lambda_{1}$ and $\lambda_{2}$ may be equal.) Let $k_{n}$ be defined by $k_{0}=0, k_{1}=1$ and for $n \geq 2$
$$
k_{n}=\sum_{i=1}^{n} \lambda_{1}^{n-i} \lambda_{2}^{i-1} .
$$
Prove that
$$
k_{n+1}=\left(\lambda_{1}+\lambda_{2}\right) k_{n}-\lambda_{1} \lambda_{2} k_{n-1},
$$
if $n \geq 1$. Also prove that
$$
k_{n}= \begin{cases}\left(\lambda_{1}^{n}-\lambda_{2}^{n}\right) /\left(\lambda_{1}-\lambda_{2}\right) & \text { if } \lambda_{1} \neq \lambda_{2} \ n \lambda_{1}^{n-1} & \text { if } \lambda_{1}=\lambda_{2}\end{cases}
$$
Use mathematical induction to prove that if $n \geq 1$,
$$
A^{n}=k_{n} A-\lambda_{1} \lambda_{2} k_{n-1} I_{2},
$$
[Hint: Use the equation $A^{2}=(a+d) A-(a d-b c) I_{2}$.]
