#### Chapter 7 Differential Calculus

Section 7.1 Derivative of a Function

# 7.1.4 Exercises

##### Exercise 7.1.5
Using the difference quotient, calculate the derivative of $f:ℝ\to ℝ$, $x\to f\left(x\right):=4-{x}^{2}$ at the points ${x}_{1}=-2$ and ${x}_{2}=1$.
1. The difference quotient of $f$ at the point ${x}_{1}=-2$ is
and has for $x\to -2$ the limit $f\text{'}\left(-2\right)=$ .
2. The difference quotient of $f$ at the point ${x}_{2}=1$ is
and has for $x\to 1$ the limit $f\text{'}\left(1\right)=$ .

##### Exercise 7.1.6
Explain why the functions
1. $f:\left[-3;\infty \left[\text{}\to ℝ$ with $f\left(x\right):=\sqrt{x+3}$ at ${x}_{0}=-3$ and
2. $g:ℝ\to ℝ$ with $g\left(x\right):=6·|2x-10|$ at ${x}_{0}=5$
are not differentiable.
1. The derivative of the function $f$ at the point ${x}_{0}=-3$ does not exist since the difference quotient
does not converge for $h\to 0$.
2. The derivative of the function $g$ at the point ${x}_{0}=5$ does not exist since the difference quotient for $h<0$ has the value
and for $h>0$ has the value
. Thus, the limit for $h\to 0$ does not exist.