微积分作业代写calclulus代考| The derivative

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微积分作业代写calclulus代考|Example 2.1:

Figure 2.1 A function not everywhere differentiable.
For the function shown in Figure 2.1, (i) and (ii) are satisfied everywhere. At $x=0$, however, although the left and right limits of (iii) exist, they are not equal, implying that no derivative exists there. Everywhere else (iii) is satisfied.

微积分作业代写calclulus代考|Example 2.2:

Figure 2.2 Another function not everywhere differentiable.
For the function shown in Figure $2.2$, the only problem appears at $x=0$. Conditions (i) and (ii) are satisfied, but in the case of condition (iii) we have that
\begin{aligned} &\lim {h \rightarrow 0^{-}} \frac{f(0+h)-f(0)}{h}=\lim {h \rightarrow 0^{-}} \frac{(1-h)-1}{h}=-1 \ &\lim {h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}=\lim {h \rightarrow 0^{+}} \frac{\left(1+h^{3 / 2}\right)-1}{h}=\lim _{h \rightarrow 0^{+}} h^{1 / 2}=0 \end{aligned}
That is, the left limit is not equal to the right limit, implying that no derivative exists there.

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Differentiation
Point (ii) is the definition of continuity at $x_{0}$ (Section 1.C). A function for which any of the equalities is not satisfied is said to be discontinuous at $x_{0}$. What we are now saying is that continuity is a necessary but not a sufficient condition for differentiability. Functions for which (iii) is not satisfied at any point, $x_{0}$, such as those of the foregoing examples, are said to be singular at that point.

Now let us apply what we have learnt for a function of one variable to the case of a function of two variables. The most obvious analogous expression of a limit generalized to some function $f: \mathbb{R}^{2} \longrightarrow \mathbb{R}$ of two variables is:
$$\lim {P{1} \rightarrow P_{0}} \frac{f\left(x_{0}, y_{0}\right)-f\left(x_{1}, y_{1}\right)}{\sqrt{\left(x_{1}-x_{0}\right)^{2}+\left(y_{1}-y_{0}\right)^{2}}} .$$
If this limit exists, should we call it “the” derivative of $f$ ? Alongside this question we also need to ask what are the generalizations of criteria (i)-(iii) to $\mathbb{R}^{2}$ (or $\left.\mathbb{R}^{3}, \ldots, \mathbb{R}^{n}\right)$ ?

The graphical foundation for the limit expression (2.1) is shown in Figure 2.3. The things to note are, firstly, the graph of $f$ is suspended in 3D; secondly, the domain $D_{f}$ lies in the $x y$-plane; thirdly, the points $P_{0}$ and $P_{1}$ in $D_{f}$ give rise to values $z_{0}$ and $z_{1}$, respectively; and finally, the line in the domain joining $P_{0}$ and $P_{1}$ traces out the black curve in the graph of $f$.

微积分作业代写calclulus代考|Example 2.2:

\begin{aligned} &\lim {h \rightarrow 0^{-}} \frac{f(0+ h)-f(0)}{h}=\lim {h \rightarrow 0^{-}} \frac{(1-h)-1}{h}=-1 \ &\lim {h \rightarrow 0 ^{+}} \frac{f(0+h)-f(0)}{h}=\lim {h \rightarrow 0^{+}} \frac{\left(1+h^{3 / 2 }\right)-1}{h}=\lim _{h \rightarrow 0^{+}} h^{1 / 2}=0 \end{aligned}

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$$\lim {P {1} \rightarrow P_{0}} \frac{f\left(x_{0}, y_{0}\right)-f\left(x_{1}, y_{1}\right)}{\sqrt{\left(x_{1}-x_{0}\right)^{2}+\left(y_{1}-y_{0}\right)^{2 }}} 。$$