这是一份southampton南安普敦大学 MATH1007W1-01作业代写的成功案例
Since $f(x)$ possesses a finite derivative at the point $x$, say $a$,
$$
\lim _{\delta x \rightarrow 0} \frac{f(x+\delta x)-f(x)}{\delta x}=a
$$
and so,
$$
|f(x+\delta x)-f(x)|=|\delta x|(|a|+\varepsilon)
$$
where $\varepsilon \rightarrow 0$ as $\delta x \rightarrow 0$. Since $a$ is finite
$$
\mid f(x)+\delta x)-f(x) \mid \rightarrow 0 \text { as }|\delta x| \rightarrow 0
$$
Hence $f(x)$ is continuous at the point $x$.
(2) If $f(x)$ is a constant, its derivative is zero, since
Differentials
$$
\lim {\delta x \rightarrow 0} \frac{f(x+\delta x)-f(x)}{\delta x}=\lim \frac{0}{\delta x}=0 $$ (3) If $f(x)=x$, its derivative is unity, since $$ \lim {\delta x \rightarrow 0} \frac{(x+\delta x)-x}{\delta x}=\lim \frac{\delta x}{\delta x}=1
$$
MATH1007W1-01 COURSE NOTES ：
If $y=\sin ^{-1} x$ where $-1<x<1$ then $x=\sin y$ and is increasing steadily in the range $-\pi / 2<y<\pi / 2$. Since $d x / d y=\cos y$, it follows
$$
\frac{d y}{d x}=\frac{1}{\cos y}
$$
However, in the range $-\pi / 2<y<\pi / 2, \cos y$ is positive and given by
$$
\begin{gathered}
\cos y=\left(1-\sin ^{2} y\right)^{1 / 2}=\left(1-x^{2}\right)^{1 / 2} \
\frac{d y}{d x}=\frac{1}{\left(1-x^{2}\right)^{1 / 2}}
\end{gathered}
$$