## The Definition of a Derivative

Read this section to understand the definition of a derivative. Work through practice problems 1-8.

Practice 1: The graph of $\mathrm{f}(\mathrm{x})=7 \mathrm{x}$ is a line through the origin. The slope of the line is $7$.

For all $x, m_{\tan }=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{7(x+h)-7 x}{h}=\lim\limits_{h \rightarrow 0} \frac{7 h}{h}=\lim\limits_{h \rightarrow 0} 7=7$

Practice 2: $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$ so $\mathrm{f}(\mathrm{x}+\mathrm{h})=(\mathrm{x}+\mathrm{h})^{3}=\mathrm{x}^{3}+3 \mathrm{x}^{2} \mathrm{~h}+3 \mathrm{xh}^{2}+\mathrm{h}^{3}$

\begin{aligned}\frac{d y}{\mathrm{dx}}=& \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{\left\{x^{3}+3 x^{2} h+3 x h^{2}+h^{3}\right\}-x^{n}}{h} \\&=\lim _{h \rightarrow 0} \frac{3 x^{2} h+3 x h^{2}+h^{3}}{h}=\lim\limits_{h \rightarrow 0}\left(3 x^{2}+3 x h+h^{2}\right)=3 \mathbf{x}^{2}\end{aligned}

At the point $(2,8)$, the slope of the tangent line is $3(2)^2 = 12$ so the equation of the tangent line is $y – 8 = 12(x – 2)$ or $y = 12x –16$.

Practice 3: $\mathbf{D}\left(\mathrm{x}^{4}\right)=4 \mathrm{x}^{3}, \mathbf{D}\left(\mathrm{x}^{5}\right)=5 \mathrm{x}^{4}, \mathbf{D}\left(\mathrm{x}^{43}\right)=43 \mathrm{x}^{42}, \mathrm{D}\left(\mathrm{x}^{1 / 2}\right)=\frac{1}{2} \mathrm{x}^{-1 / 2}, \mathrm{D}\left(\mathrm{x}^{\pi}\right)=\pi \mathrm{x}^{\pi-1}$

Practice 4:

\begin{aligned}&D(\cos (x))=\lim\limits_{h \rightarrow 0} \frac{\cos (x+h)-\cos (x)}{h}=\lim\limits_{h \rightarrow 0} \frac{\cos (x) \cos (h)-\sin (x) \sin (h)-\cos (x)}{h} \\&=\lim\limits_{h \rightarrow 0} \cos (x) \frac{\cos (h)-1}{h}-\sin (x) \frac{\sin (h)}{h} \longrightarrow \cos (x) \cdot(0)-\sin (x) \cdot(1)=-\sin (x)\end{aligned}

Practice 5: See Fig. 16 for the graphs of $\mathrm{y}=|\mathrm{x}-2|$ and $\mathrm{y}=|2 \mathrm{x}|$

\begin{aligned}&\mathbf{D}(|x-2|)= \begin{cases}+1 & \text { if } x > 2 \\\text { undefined } & \text { if } x = 2 \\-1 & \text { if } x < 2\end{cases} \\&D(|2 x|)= \begin{cases}+2 & \text { if } x > 0 \\\text { undefined } & \text { if } x = 0 \\-2 & \text { if } x < 0\end{cases}\end{aligned}

Practice 6:

$h(t) = sin(t)$ so $h(1) = sin(1) ≈ 0.84$ feet,
$v(t) = cos(t)$ so $v(1) =cos(1) ≈ 0.54$ feet/second.
$a(t) = –sin(t)$ so $a(1) = –sin(1) ≈ –0.84$ feet/second2.

Practice 7: $\quad \mathbf{D}\left(\mathrm{x}^{5}\right)=5 \mathrm{x}^{4}, \quad \frac{\mathbf{d} \mathrm{x}^{2}}{\mathrm{dx}}=2 \mathrm{x}^{1}=2 \mathrm{x}, \frac{\mathbf{d} \mathrm{x}^{100}}{\mathrm{dx}}=100 \mathrm{x}^{99}, \frac{\mathbf{d} \mathrm{t}^{31}}{\mathrm{dt}}=31 \mathrm{t}^{30}$,

$\mathbf{D}\left(x^{0}\right)=0 x^{-1}=0 \text { or } \mathbf{D}\left(x^{0}\right)=D(1)=0$

Practice 8:

$\mathbf{D}\left(\mathrm{x}^{3 / 2}\right)=\frac{3}{2} \mathrm{x}^{1 / 2}, \quad \frac{\mathbf{d} \mathrm{x}^{1 / 3}}{\mathrm{dx}}=\frac{1}{3} \mathrm{x}^{-2 / 3}, \quad \mathbf{D}\left(\mathrm{x}^{-1 / 2}\right)=\frac{-1}{2} \mathrm{x}^{-3 / 2}, \quad \frac{\mathbf{d} \mathrm{t}^{\pi}}{\mathrm{dt}}=\pi \mathrm{t}^{\pi-1}$