112 lines
3.1 KiB
Markdown
112 lines
3.1 KiB
Markdown
# Math4121 Lecture 1
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## Chapter 5: Differentiation
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### The derivative of a real function
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#### Definition 5.1
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Let $f$ be a real-valued function on an interval $[a,b]$ ($f: [a,b] \to \mathbb{R}$).
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We say that $f$ is _differentiable_ at a point $x\in [a,b]$ if the limit
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$$
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\lim_{t\to x} \frac{f(t)-f(x)}{t-x}
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$$
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exists.
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Then we defined the derivative of $f$, $f'$, a function whose domain is the set of all $x\in [a,b]$ at which $f$ is differentiable, by
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$$
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f'(x) = \lim_{t\to x} \frac{f(t)-f(x)}{t-x}
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$$
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#### Theorem 5.2
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Let $f:[a,b]\to \mathbb{R}$. If $f$ is differentiable at $x\in [a,b]$, then $f$ is continuous at $x$.
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Proof:
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> Recall [Definition 4.5](https://notenextra.trance-0.com/Math4111/Math4111_L22#definition-45)
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>
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> $f$ is continuous at $x$ if $\forall \epsilon > 0, \exists \delta > 0$ such that if $|t-x| < \delta$, then $|f(t)-f(x)| < \epsilon$.
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>
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> Whenever you see a limit, you should think of this definition.
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We need to show that $\lim_{t\to x} f(t) = f(x)$.
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Equivalently, we need to show that
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$$
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\lim_{t\to x} (f(t)-f(x)) = 0
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$$
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So for $t\ne x$, since $f$ is differentiable at $x$, we have
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$$
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\begin{aligned}
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\lim_{t\to x} (f(t)-f(x)) &= \lim_{t\to x} \left(\frac{f(t)-f(x)}{t-x}\right)(t-x) \\
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&= \lim_{t\to x} \left(\frac{f(t)-f(x)}{t-x}\right) \lim_{t\to x} (t-x) \\
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&= f'(x) \cdot 0 \\
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&= 0
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\end{aligned}
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$$
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Therefore, differentiable is a stronger condition than continuous.
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> There exists some function that is continuous but not differentiable.
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>
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> For example, $f(x) = |x|$ is continuous at $x=0$, but not differentiable at $x=0$.
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>
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> We can see that the left-hand limit and the right-hand limit are not the same.
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>
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> $$ \lim_{t\to 0^-} \frac{|t|-|0|}{t-0} = -1 \quad \text{and} \quad \lim_{t\to 0^+} \frac{|t|-|0|}{t-0} = 1 $$
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>
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> Therefore, the limit does not exist. for $f(x) = |x|$ at $x=0$.
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#### Theorem 5.3
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Suppose $f$ is differentiable at $x\in [a,b]$ and $g$ is differentiable at a point $x\in [a,b]$. Then $f+g$, $fg$ and $f/g$ are differentiable at $x$, and
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(a) $(f+g)'(x) = f'(x) + g'(x)$
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(b) $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$
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(c) $\left(\frac{f}{g}\right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$, provided $g(x)\ne 0$
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Proof:
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Since the limit of product is the product of the limits, we can use the definition of the derivative to prove the theorem.
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(a)
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$$
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\begin{aligned}
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(f+g)'(x) &= \lim_{t\to x} \frac{(f+g)(t)-(f+g)(x)}{t-x} \\
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&= \lim_{t\to x} \frac{f(t)-f(x)}{t-x} + \lim_{t\to x} \frac{g(t)-g(x)}{t-x} \\
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&= f'(x) + g'(x)
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\end{aligned}
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$$
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(b)
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Since $f$ is differentiable at $x$, we have $\lim_{t\to x} f(t) = f(x)$.
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$$
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\begin{aligned}
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(fg)'(x) &= \lim_{t\to x} \left(\frac{f(t)g(t)-f(x)g(x)}{t-x}\right) \\
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&= \lim_{t\to x} \left(f(t)\frac{g(t)-g(x)}{t-x} + g(x)\frac{f(t)-f(x)}{t-x}\right) \\
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&= f(t) \lim_{t\to x} \frac{g(t)-g(x)}{t-x} + g(x) \lim_{t\to x} \frac{f(t)-f(x)}{t-x} \\
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&= f(x)g'(x) + g(x)f'(x)
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\end{aligned}
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$$
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(c)
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$$
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\begin{aligned}
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\left(\frac{f}{g}\right)'(x) &= \lim_{t\to x}\left(\frac{f(t)g(x)}{g(t)g(x)} - \frac{f(x)g(x)}{g(t)g(x)}\right) \\
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&= \frac{1}{g(t)g(x)}\left(\lim_{t\to x} (f(t)g(x)-f(x)g(t))\right) \\
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\end{aligned}
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$$
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