Derivatives, Properties, and Formulas

Practice 1: $\mathrm{f}$ is continuous at $\mathrm{x}=-1,0,2,4,6$, and $7 . \mathrm{f}$ is differentiable at $\mathrm{x}=-1,2,4$, and $7$

Practice 2: $\quad \mathrm{f}(\mathrm{x})=6 \mathrm{x}+8$ and $\mathrm{g}(\mathrm{x})=2$ so $\mathrm{D}(\mathrm{f}(\mathrm{x}))=6$ and $\mathrm{D}(\mathrm{g}(\mathrm{x}))=0$.

$\begin{aligned} &D(3 f(x))=3 D(f(x))=3(6)=18, \quad D(x)+g(x))=D(f(x))+D(g(x))=6+0=6 \\ &D(f(x)-g(x))=D(f(x))-D(g(x))=6-0=6 \\ &D(f(x) g(x))=f(x) D(g(x))+g(x) D(f(x))=(6 x+8)(0)+(2)(6)=12 \\ &D(f(x) / g(x))=\frac{g(x) D(f(x))-f(x) D(g(x))}{(g(x))^{2}}=\frac{(2)(6)-(6 x+8)(0)}{2^{2}}=\frac{12}{4}=3 \end{aligned}$

Practice 3: $\quad D\left(x^{3}-5 \sin (x)\right)=D\left(x^{3}\right)-5 D(\sin (x))=3 x^{2}-5 \cos (x)$

$\quad  \frac{\mathbf{d}}{\mathbf{d x}}\left(\sin (x)-4 x^{3}\right)=\frac{\mathbf{d}}{d x} \sin (x)-4 \frac{d}{d x} x^{3}=\cos (x)-12 \mathbf{x}^{2}$

Practice 4:

Practice 5: 

$\begin{align*} \begin{aligned} D(f(x)+g(x))=& \lim\limits_{h \rightarrow 0} \frac{\{f(x+h)+g(x+h)\}-\{f(x)+g(x)\}}{h} \\ &=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)+g(x+h)-g(x)}{h} \\ &=\left(\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right)+\left(\lim\limits_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}\right)=\mathbf{D}(\mathrm{f}(\mathrm{x}))+\mathbf{D}(\mathrm{g}(\mathrm{x})) \end{aligned} \end{align*}$

Practice 6:

$\begin{array}{l|l|l|l|c|c|c|c} x & \mathrm{f}(\mathrm{x}) & \mathrm{f}^{\prime}(\mathrm{x}) & \mathrm{g}(\mathrm{x}) & \mathrm{g}^{\prime}(\mathrm{x}) & \mathrm{D}(\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})) & \mathrm{D}(\mathrm{f}(\mathrm{x}) / \mathrm{g}(\mathrm{x})) & \mathrm{D}(\mathrm{g}(\mathrm{x}) / \mathrm{f}(\mathrm{x})) \\ \hline 0 & 3 & -2 & -4 & 3 & 3 \cdot 3+(-4)(-2)=\mathbf{1} 7 & \frac{-4(-2)-(3)(3)}{(-4)^{2}}=\frac{-\mathbf{1}}{\mathbf{1 6}} & \frac{(3)(3)-(-4)(-2)}{3^{2}}=\frac{1}{\mathbf{9}} \\ 1 & 2 & -1 & 1 & 0 & 2 \cdot 0+1(-1)=-\mathbf{1} & \frac{1(-1)-(2)(0)}{1^{2}}=-\mathbf{1} & \frac{2(0)-1(-1)}{2^{2}}=\frac{1}{4} \\ 2 & 4 & 2 & 3 & 1 & 4 \cdot 1+3 \cdot 2=\mathbf{1 0} & \frac{3(2)-(4)(1)}{3^{2}}=\frac{2}{9} & \frac{4(1)-3(2)}{4^{2}}=\frac{-\mathbf{2}}{\mathbf{1 6}} \end{array}$

Practice 7:

$\left.\mathbf{D}\left(\left(x^{2}+1\right)(7 x-3)\right)=\left(x^{2}+1\right) \mathbf{D}(7 x-3)+(7 x-3)\right)\left(x^{2}+1\right)=\left(x^{2}+1\right)(7)+(7 x-3)(2 x)=21 x^{2}-6 x+7$

or $D\left(\left(x^{2}+1\right)(7 x-3)\right)=D\left(7 x^{3}-3 x^{2}+7 x\right)=21 x^{2}-6 x+7$

$\begin{aligned}&\begin{array}{l}\frac{\text { d }}{\text { dt }}\left(\frac{3 t-2}{5 t+1}\right)=\frac{(5 t+1) D(3 t-2)-(3 t-2) D(5 t+1)}{(5 t+1)^{2}}=\frac{(5 t+1)(3)-(3 t-2)(5)}{(5 t+1)^{2}}=\frac{13}{(5 t+1)^{2}} \\ D\left(\frac{\cos (x)}{x}\right)=\frac{x D(\cos (x))-\cos (x) D(x)}{(x)^{2}}=\frac{x(-\sin (x))-\cos (x)(1)}{x^{2}}=\frac{-x \cdot \sin (x)-\cos (x)}{x^{2}} \end{array} \end{aligned}$

Practice 8: (a) difference rule (b) product rule (c) difference rule (d) quotient rule

Practice 9:

$\begin{aligned} &\mathbf{D}\left(\frac{x^{2}-5}{\sin (x)}\right)=\frac{\sin (x) D\left(x^{2}-5\right)-\left(x^{2}-5\right) D(\sin (x))}{(\sin (x))^{2}}=\frac{\sin (x)(2 x)-\left(x^{2}-5\right) \cos (x)}{\sin ^{2}(x)} \\ &\frac{\mathbf{d}}{\mathbf{d t}}\left(\frac{t^{2}-5}{t \cdot \sin (t)}\right)=\frac{t \cdot \sin (t) D\left(t^{2}-5\right)-\left(t^{2}-5\right) \mathbf{D}(t \cdot \sin (t))}{(t \cdot \sin (t))^{2}}=\frac{t \cdot \sin (t)(2 t)-\left(t^{2}-5\right)\{\mathbf{t} \cdot \cos (t)+\sin (\mathbf{t})\}}{t^{2} \cdot \sin ^{2}(t)} \end{aligned}$

Practice 10:

(a) $\quad \mathrm{h}(5)=5+\frac{\sin (5)}{1+5} \approx 4.84$ ft. $\mathrm{v}(5)=\mathrm{h}^{\prime}(5)=\frac{(1+5) \cos (5)-\sin (5)}{(1+5)^{2}} \approx 0.074 \mathrm{ft} / \mathrm{sec} .$

"long time": $\quad \lim\limits_{t \rightarrow \infty} h(t)=\lim\limits_{t \rightarrow \infty} 5+\frac{\sin (t)}{1+t}=5$ feet.

$\lim\limits_{t \rightarrow \infty} h^{\prime}(t)=\lim\limits_{t \rightarrow \infty} \frac{(1+t) \cos (t)-\sin (t)}{(1+t)^{2}}=\lim\limits_{t \rightarrow \infty}\left\{\frac{\cos (t)}{1+t}-\frac{\sin (t)}{(1+t)^{2}}\right\}=0 \mathbf{f t} / \mathbf{s e c}$

Practice 11:

$f^{\prime}(x)=2 x-10 . f^{\prime}(x)=0$ when $2 x-10=0$ so when $x=5$.

$g^{\prime}(x)=3 x^{2}-12 . g^{\prime}(x)=0$ when $3 x^{2}-12=0$ so $x^{2}=4$ and $x=-2,+2$