If not stated otherwise, we assume that the metric spaces \mathbb{R} and \mathbb{Q} are quipped with the Euclidean metric d(x,y)=|x-y|.

Exercise: Use definition to determine the derivative of the function

f(x)=\frac{1}{\sqrt{x}}, x>0

Proof. Using the definition, we can calculate the derivative

\begin{aligned}\lim_{t\rightarrow x}\frac{f(t)-f(x)}{t-x}&= \lim_{t\rightarrow x}\frac{1/\sqrt{t}-1/\sqrt{x}}{t-x}\\&= \lim_{t\rightarrow x}-\frac{1}{\sqrt{tx}(\sqrt{t}+\sqrt{x})}\\&= -\frac{1}{2x\sqrt{x}}\end{aligned}

Thus f'(x)=-\frac{1}{2x\sqrt{x}}

Exercise: Let f be defined for all x\in\mathbb{R}, and suppose that

|f(x)-f(y)|\leq |x-y|^2

for all x,y\in\mathbb{R}. Prove that f'(x)=0 for all x\in\mathbb{R}. Prove that f'(x)=0 for all x\in\mathbb{R}. Hence f is a constant function.

Proof. Whenever x\neq y, |f(x)-f(y)|/|x-y|\leq|x-y|. So \lim_{y\rightarrow x}|f(x)-y(y)|/|x-y|=0, which implies that f'(x)=0 for all x\in\mathbb{R}.

Exercise: Assume f:(a,b)\rightarrow \mathbb{R} is injective, and let g denote its left inverse function. That is, for all x\in(a,b), g(f(x))=x.

Further, assume that

- f is differentiable at a point x\in(a,b) with f'(x)\neq 0. Prove that if g is continuous at the point f(x), then g is differentiable at f(x), and g'(f(x))=1/f'(x). Hint: Use Theorem 4.2.
- f is continuous on a,b and differentiable at a point x\in(a,b) with f'(x)\neq 0. Prove that g is differentiable at the point f(x), and that g'(f(x))=1/f'(x). Hint: Use Theorem 4.2. Also you may use, without proof, the fact that g is continuous at f(x).

Proof. Let Y be the range of f. Pick any sequence in Y, say \{y_n\}, whose limit is y=f(x).

We want to show that \lim_{n\rightarrow\infty}\frac{g(y_n)-g(y)}{y_n-y} exists, and it’s equal to 1/f'(x).

As y_n\in Y and f is injective, let x_n be the unique number in (a,b) such that f(x_n)=y_n, hence x_n=g(f(x_n))=g(y_n).

As g is continuouss at y, \lim_{n\rightarrow\infty}x_n=g(\lim_{n\rightarrow\infty}y_n)=g(y)=g(f(x))=x. Thus \frac{g(y_n)-g(y)}{y_n-y}=\frac{x_n-x}{f(x_n)-f(x)}, whence g'(f(x)) exists as f'(x)\neq 0, and it’s equal to 1/f'(x).

Exercise: Let h be the real-valued function defined by

h(x)=\bigg\{{\begin{array}{cc}f(x)&\text{if }a\leq x\leq b\\g(x)&\text{if }b\leq x\leq c\end{array}}

where f:[a,b]\rightarrow\mathbb{R} and g:[b,c]\rightarrow\mathbb{R} are differentiable and f(b)=g(b). Prove that h is differentiable on [a,c] if and only if f'(b)=g'(b).

Proof. (Sketch) Easy to see that h is differentiable on [a,c] if and only if h is differentiable at b, which is equivalent to say f'(b)=g'(b) by the definition of differentiability.