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Find higher derivatives of rational and radical functions

{: \begin{align*} &f(x)=3x-5x^{2}-x^{3},\ &g(x)=6x^{2}-4x \end{align*} :}

Given the following table of values, find

h^{\prime}(2)

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\begin{array}{l} h(x)=\frac{4}{x^{3}}+\frac{f(x)}{g(x)} \text {. } \\ \begin{array}{rrrrr} x & f(x) & f^{\prime}(x) & g(x) & g^{\prime}(x) \\ \hline 2 & 3 & -3 & 3 & 1 \end{array} \\ \end{array}

x

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\begin{array}{l} 3^{10}=3^{7} \cdot x^{3} \\ x=\square \end{array}

2

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Right Triangles

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\quad

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Find the measure of each side indicated. Round to the nearest tenth.

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1

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\begin{align*}\(\newline\sin 52 &= \frac{13}{x} (\newline\)\frac{x}{\cancel{7881}}(13) &= \frac{x}{\cancel{x}} \times x (\newline\)\frac{7881}{7881} &= \frac{13}{7881} (\newline\)x &= 16.49

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Solve by completing the square

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\begin{array}{l} \text { olve by completung } \\ 5 x^{2}+3 x-4=0 \end{array}

The second derivative of a function

f

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f^{\prime \prime}(x)=\sin \left(x^{2}\right)+\cos (2 x)-\frac{1}{2} \text {. }

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How many points of inflection does the graph of

f

have on the interval

1<x<4

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Use a graphing calculator.

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1

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The derivative of a function

f

f^{\prime}(x)=e^{x}-x^{3}

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On which intervals is the graph of

f

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Use a graphing calculator.

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1

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[1.857,4.536]

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(-\infty, 1.857]

[4.536, \infty)

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(-\infty,-0.459]

[0.91,3.733]

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[-0.459,0.91]

[3.733, \infty)

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(E) All real numbers

The second derivative of a function

h

h^{\prime \prime}(x)=\frac{\sqrt[3]{x}}{e^{x}}+1

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Where does the graph of

h

have a point of inflection?

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Use a graphing calculator.

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1

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x=-0.35

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x=0

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x=0.33

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x=1.497

\begin{array}{l}f^{\prime}(x)=-\frac{6}{x^{4}} \text { and } f(1)=-4 . \\ f(-1)=\end{array}

\begin{array}{l}f^{\prime}(x)=\frac{16}{x^{2}} \text { and } f(-2)=0 . \\ f(4)=\square\end{array}

\begin{array}{l}f^{\prime}(x)=\frac{81}{x^{4}} \text { and } f(-1)=18 . \\ f(3)=\end{array}

\begin{array}{l}f^{\prime}(x)=\frac{24}{x^{3}} \text { and } f(2)=12 . \\ f(-1)=\end{array}

\begin{array}{l}G(x)=(x+3)^{2} \\ g(x)=G^{\prime}(x) \\ \int_{-1}^{7} g(x) d x=\end{array}

F(x)=4 x^{2}-5 x

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\begin{array}{l}f(x)=F^{\prime}(x) \\ \int_{1}^{4} f(x) d x=\end{array}

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\begin{array}{l} x^{2}+5 x+6 \\ P=6 x^{2} \end{array}

\begin{array}{l}\int u \cdot v^{\prime} d x \\ =u \cdot v-\int u^{\prime} v d x\end{array}

7

. If the minute hand of a clock is

3

5

inches long and its tip rotates through a distance of

8

inches, then which of the following is closest to the angle that it rotates?

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1

131^{\circ}

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3

267^{\circ}

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\begin{array}{l} \theta=\frac{5}{r} \\ \frac{8}{3.5}\left(\frac{180}{\pi}\right) \end{array}

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2

174^{\circ}

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4

314^{\circ}

SD = \sqrt[i]{\left(\frac{\sum f x^{2}}{n}\right)-\left(\frac{kfx^{1}}{n}\right)}

SD = \sqrt[7]{\left(\frac{261}{90}\right)-\left(\frac{-128}{90}\right)^{2}}

=7\sqrt{2,9-(-1,422)^{2}}

=7\sqrt{2,9-2,0224}

=7

n

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\frac{12}{5}=\frac{n}{8}

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n =\square

\begin{array}{l}f(x)=-5 x-9 \\ g(x)=2 x^{2} \\ g(f(0))=?\end{array}

\begin{array}{c}(-10)-(14)= \\ (3)-(18)= \\ (19)+(-19)= \\ (-27)-(-25)= \\ (38)-(20)= \\ (10)+(-16)= \\ (-21)+(7)= \\ (-11)+(-13)= \\ (21)+(16)= \\ (-6)+(-1)=\end{array}

\begin{array}{l}x^{2}+x-72=0 \\ (x+\square)^{2}=\square\end{array}

Find the missing term.

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(x+2)(x+5)

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x^{2}+5 x+2 x+(2) \cdot(5)

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x^{2}\_\_\_\_\_+10

\begin{array}{l}y=m x+b \\ y=-\frac{5}{4} x+b \\ -10=-\frac{5}{4}(15)+b \\ -10=-\frac{25}{4}+b \\ -10+\frac{25}{4}=b, \quad b=-\frac{15}{4} \\ y=-\frac{5}{4} x-\frac{15}{4}\end{array}

z_{1} z_{2}

. Write the answer in standard form.

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\begin{array}{l} z_{1}=7 \operatorname{cis} \frac{\pi}{3} \\ z_{2}=2 \operatorname{cis} \frac{11 \pi}{6} \end{array}

\left\{\begin{array}{l} b(1)=-500 \\ b(n)=b(n-1) \cdot \frac{4}{5} \end{array}\right.

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3^{\text {rd }}

term in the sequence?

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\square

\begin{array}{l}10^{x+2 y}=5 \\ 10^{3 x+4 y}=50\end{array}

\begin{array}{l}\text { j) }-\frac{3}{3+4 y}=\frac{1}{x-z} \\ \frac{3}{x+y-1}=-z^{-1} \\ \frac{6 z}{5 y+8}=-1 /: 5 y+8\end{array}

y=4 x^{2}+9 x

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Опиши кроки розв'язку:

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1

f(x)=

x^{2}+

x

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2

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\begin{array}{l} f(x+\Delta x)=\square x^{2}+\square x . \\ \Delta x+\square \cdot(\Delta x)^{2}+\square x+ \\ \square \Delta \end{array}

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3

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\begin{array}{l} \Delta y=\square x \cdot \Delta x+ \\ \square(\Delta x)^{2}+\square \cdot \Delta x \end{array}

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4

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\frac{\Delta y}{\Delta x}=\square x+\square \cdot \Delta x+

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5

f^{\prime}(x)=

x+

\begin{array}{l}y=4-x^{2} \\ y=3 x\end{array}

\frac{2}{x+1}+\frac{x}{x-1}=\frac{2}{x^{2}-1}

\begin{array}{l}P(x)=x^{4}-2 x^{2}-8 \\ d(x)=x+2\end{array}

852

. Поставьте вместо знака

*

такой одночлен, чтобы трёхчлен но было представить в виде квадрата двучлена:

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*+56 a+49

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36-12 x+*

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25 a^{2}+*+\frac{1}{4} b^{2}

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0,01 b^{2}+*+100 c^{2}

34

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\begin{aligned} \sqrt[4]{25 m^{2}} & =\sqrt[2]{25 m} \\ & =5 m \end{aligned}

T(x) = \frac{e^{x^2 + 8x}}{\sqrt{x^4 + 7}},\quad T'(x) =

\begin{cases} f(x)=2x^{4}+x^{3}-11x^{2}+11x-3, \ 2, \frac{3}{2} \end{cases}

Complete the recursive formula of the arithmetic sequence

\{:[-15, -11, -7, -3, \dots], [c(1)=\square], [c(n)=c(n-1)+\square]:\}

Use the chain rule to find the indicated partial derivative

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\begin{array}{l} z=x^{4}+x^{2} y, \quad x=s+2 t-u, \quad y=s t^{2} \\ \frac{\partial z}{\partial s}, \frac{\partial z}{\partial t}, \frac{\partial z}{\partial u} \text { when } s=3, t=1, u=2 \end{array}

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\frac{\partial z}{\partial s}=

\begin{array}{r}4 x_{2}+x_{3}=0 \\ x_{1}+x_{2}+x_{3}=0 \\ 2 x_{1}+4 x_{2}+3 x_{3}=0\end{array}

\begin{array}{l} =x^{2}+4 \\ =(g+h)(x) \end{array}

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9

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\begin{array}{l} h(x)=x^{2}+x \\ g(x)=3 x+5 \\ \text { Find }(h \cdot g)(x) \end{array}

\begin{array}{c}x^{2}-6 x+11=y \\ x=y+1\end{array}

\begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}

n

\frac{12}{5}=\frac{n}{8}

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n = \Box

Solve the equation. \begin{align*} \frac{b}{6} &= 3, \ b &= \Box \end{align*}

Solve the equation.

\begin{align*} \frac{b}{6} &= 3 \\ b &= \Box \end{align*}

n

\begin{array}{l}\frac{8}{3}=\frac{12}{n} \\ n=\square\end{array}

\begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}

\begin{array}{l}f^{\prime}(x)=5 e^{x} \text { and } f(7)=40+5 e^{7} . \\ f(0)=\square\end{array}

\begin{array}{l}f^{\prime}(x)=-5 e^{x} \text { and } f(3)=22-5 e^{3} . \\ f(0)=\square\end{array}

\begin{array}{l}f^{\prime}(x)=-27 e^{x} \text { and } f(6)=36-27 e^{6} . \\ f(0)=\square\end{array}