Resources

Resources

Find derivatives of using multiple formulae

For the following equation, find

\frac{d y}{d x}

\newline

y=3 x^{5}+9 x^{4}-7 x^{3}+3 x

\newline

\frac{d y}{d x}=

\frac{d}{dx}(-4x^{4}-x^{-2}-5)

\frac{dy}{dx}\left(x\right)\left(\frac{1}{\left(3x-2\right)^{9}}\right)

\frac{dy}{dx}\left(x\right)\left(\frac{1}{\left(3x-2\right)^{6}}\right)

The following are all angle measures, in degrees, whose cosine is

-1

\newline

Which is the principal value of

\newline

\arccos(-1)

\newline

1

\newline

-540^{\circ}

\newline

-180^{\circ}

\newline

180^{\circ}

\newline

540^{\circ}

Which of the following is equivalent to

\newline

(2^{x})^{3}

\newline

1

\newline

6^{x}

\newline

6x^{x^{3}}

\newline

8^{x}

\newline

8^{3x}

f(x)=\left(\frac{1}{4}\right)^x

, what is the value of

f(3)

\newline

1

\newline

\frac{3}{4}

\newline

\frac{1}{16}

\newline

\frac{1}{64}

\newline

\frac{1}{256}

Given the function

f(x)=-5\left(5 x^{2}+6 x-8\right)^{5}

f^{\prime}(x)

\newline

f^{\prime}(x)=

-1-x^{2}+y^{2}=y^{3}-5 y

\frac{d y}{d x}

(1,-2)

\newline

\left.\frac{d y}{d x}\right|_{(1,-2)}=

-y+y^{3}-3 y^{2}+3 x^{2}=0

\frac{d y}{d x}

(-1,3)

\newline

\left.\frac{d y}{d x}\right|_{(-1,3)}=

\frac{d}{d v}\left(v^{4}-4 \sin v\right)

\newline

\frac{d}{d y}\left(y^{4}-3 \sin y\right)

\newline

A curve is defined by the parametric equations

x(t)=10 t^{3}-5 t^{2}+6 t

y(t)=-t^{3}

\frac{d y}{d x}

\newline

A curve is defined by the parametric equations

x(t)=9 t^{3}+2 t^{2}-5 t-10

y(t)=-8 t^{2}

\frac{d y}{d x}

\newline

j=\frac{m}{c}\cdot 78

\newline

Which of the following equations correctly expresses

\newline

c

j

m

\newline

1

\newline

c=\frac{m}{j}\cdot 78

\newline

c=\frac{m}{78\cdot j}

\newline

c=\frac{j}{m}\cdot 78

\newline

c=\frac{j}{78\cdot m}

\newline

\newline

\frac{\cot^{2}x\cos^{2}x}{\cot^{2}x-\cos^{2}x}

Which of the following expressions is equivalent to

\newline

6\sqrt[3]{640f^{2}g^{9}}

\newline

1

\newline

60 fg^{4}\sqrt[3]{8g}

\newline

60g^{3}\sqrt[3]{4f^{2}}

\newline

48 fg^{4}\sqrt[3]{10 g}

\newline

24g^{3}\sqrt[3]{10f^{2}}

Which of the following expressions is equivalent to

\left(3 c-4 c^{2}+c^{3}\right)-\left(5 c^{2}+8 c^{3}-6 c\right)

\newline

1

\newline

7 c^{3}+9 c^{2}-3 c

\newline

7 c^{3}+c^{2}-3 c

\newline

-7 c^{3}+c^{2}+9 c

\newline

-7 c^{3}-9 c^{2}+9 c

Which of the following expressions is equivalent to

\left(5 x^{2}-6 x+7\right)-\left(3 x^{3}+4 x^{2}-8\right)

\newline

1

\newline

-3 x^{3}+x^{2}-6 x-1

\newline

-3 x^{3}+x^{2}-6 x+15

\newline

-2 x^{3}-6 x+15

\newline

-2 x^{3}-6 x-1

Which of the following is equivalent to

\left(8 r^{4}+4 r^{2}+2 r\right)-\left(8 r^{4}+4 r^{2}+1\right)

\newline

1

\newline

8 r^{2}+2 r+1

\newline

8 r^{2}+3 r

\newline

2 r-1

\newline

r

Hiroki and Mapiya were asked to find an explicit formula for the sequence

\newline

125, 25, 5, 1, \dots

, where the first term should be

\newline

f(1)

\newline

Hiroki said the formula is

\newline

f(n) = 625 \cdot \left(\frac{1}{5}\right)^n

\newline

Mapiya said the formula is

\newline

f(n) = 125 \cdot \left(\frac{1}{5}\right)^n

\newline

Which one of them is right?

\newline

1

\newline

(A) Only Hiroki

\newline

(B) Only Mapiya

\newline

(C) Both Hiroki and Mapiya

\newline

(D) Neither Hiroki nor Mapiya

(49x^{9}-56x^{8}+3x^{7}-49x^{6})/(-7x^{6})

3x^{2}z^{2}-6xz^{2}-24z^{2}

x^{16}

(4x^{3}-3x^{2}+5x)^{7}

\frac{45x^{3}+63x^{2}-72x-36}{9x}

y=a^{2}\arcsin\left(\frac{x}{a}\right)-x\sqrt{a^{2}-x^{2}}\quad(a=\text{constant})

Given the function

y=(-8 x-10)\left(-6 x^{3}-8-7 x^{-1}\right)

\frac{d y}{d x}

\newline

\frac{d y}{d x}=

Given the function

f(x)=\left(-10-2 x^{2}+5 x^{-1}\right)\left(-8-x^{2}\right)

f^{\prime}(x)

\newline

f^{\prime}(x)=

Given the function

f(x)=\left(-5-4 x^{-2}\right)\left(x^{2}+6+6 x\right)

f^{\prime}(x)

\newline

f^{\prime}(x)=

Given the function

f(x)=\left(1+3 x^{-3}+6 x^{-2}\right)\left(-8 x^{2}-6\right)

f^{\prime}(x)

\newline

f^{\prime}(x)=

Given the function

f(x)=\left(-5-4 x-9 x^{2}\right)\left(6 x^{3}+10\right)

f^{\prime}(x)

\newline

f^{\prime}(x)=

Given the function

f(x)=\left(6 x^{2}-3\right)\left(5-2 x^{2}+7 x^{-1}\right)

f^{\prime}(x)

\newline

f^{\prime}(x)=

Given the function

f(x)=\left(6+10 x^{3}\right)\left(8 x^{3}-10+8 x\right)

f^{\prime}(x)

\newline

f^{\prime}(x)=

Given the function

f(x)=\left(8 x^{3}-9 x^{2}+9\right)\left(-9+2 x^{2}\right)

f^{\prime}(x)

\newline

f^{\prime}(x)=

Given the function

f(x)=\left(-4-6 x^{3}\right)\left(-9 x^{3}-4 x^{2}+5\right)

f^{\prime}(x)

\newline

f^{\prime}(x)=

Given the function

f(x)=\left(7-8 x-10 x^{3}\right)\left(-6 x^{2}-5\right)

f^{\prime}(x)

\newline

f^{\prime}(x)=

Given the function

f(x)=\left(2 x^{3}-7\right)\left(-4 x^{3}+10-3 x\right)

f^{\prime}(x)

\newline

f^{\prime}(x)=

Given the function

f(x)=\left(-7 x^{3}-2\right)\left(5 x^{2}-6+2 x^{-3}\right)

f^{\prime}(x)

\newline

f^{\prime}(x)=

\frac{d y}{d t}=-7 y

y=3

t=0

\newline

Solve the equation.

\newline

1

\newline

y=3 e^{-7 t}

\newline

y=e^{-7 t}

\newline

y=3+e^{-7 t}

\newline

y=2+e^{-7 t}

\frac{d y}{d t}=-7 y

y=3

t=0

\newline

Solve the equation.

\newline

1

\newline

y=3 e^{-7 t}

\newline

y=3+e^{-7 t}

\newline

y=2+e^{-7 t}

\newline

y=e^{-7 t}

f(x)=\frac{3 x^{2}+1}{x+2}

\newline

f^{\prime}(-3)

\newline

1

\newline

\mathbf{1 0}

\newline

28

\newline

-28

\newline

-10

What is the value of

\frac{d}{d x}\left(\frac{x^{2}-2 x+3}{x+1}\right)

x=1

\newline

1

\newline

-2

\newline

-\frac{1}{2}

\newline

1

\newline

-1

\frac{d}{d x}\left(\frac{e^{x}}{\sqrt{x}}\right)

\newline

1

\newline

e^{x}\left(\sqrt{x}-\frac{1}{2 \sqrt{x}}\right)

\newline

2 \sqrt{x} e^{x}

\newline

e^{x}\left(\sqrt{x}-\frac{1}{2 \sqrt{x}}\right)

\newline

e^{x}-\frac{1}{2 \sqrt{x}}

y=\frac{e^{x}}{x}

\newline

\frac{d y}{d x}

\newline

1

\newline

e^{x}

\newline

\frac{e^{x}-1}{x^{2}}

\newline

\frac{e^{x-1}}{x^{2}}

\newline

\frac{e^{x}(x-1)}{x^{2}}

g(x)=\frac{\cos (x)}{\sin (x)}

\newline

g^{\prime}(x)

\newline

1

\newline

-\frac{1}{\sin ^{2}(x)}

\newline

\frac{\sin ^{2}(x)-\cos ^{2}(x)}{\sin ^{2}(x)}

\newline

\frac{1}{\sin ^{2}(x)}

\newline

\frac{\cos ^{2}(x)-\sin ^{2}(x)}{\sin ^{2}(x)}

g(x)=\frac{\sin (x)}{e^{x}}

\newline

g^{\prime}(x)

\newline

1

\newline

\cos (x)-e^{x}

\newline

e^{x}(\cos (x)-\sin (x))

\newline

\frac{\cos (x)}{e^{x}}

\newline

\frac{\cos (x)-\sin (x)}{e^{x}}

Solve the equation.

\newline

\frac{d y}{d x}=-\frac{e^{x} y^{2}}{10}

\newline

1

\newline

y=\frac{10}{e^{x}}+C

\newline

y=\frac{10}{e^{x}+C}

\newline

y=\frac{10}{C e^{x}}

\newline

y=\frac{10+C}{e^{x}}

Marcus tried to solve the differential equation

\frac{d y}{d x}=\frac{1}{3 x^{2} y^{2}}

. This is his work:

\newline

\frac{d y}{d x}=\frac{1}{3 x^{2} y^{2}}

\newline

1

\quad \int 3 y^{2} d y=\int \frac{1}{x^{2}} d x

\newline

2

\quad y^{3}=-\frac{1}{x}+C

\newline

3

\quad y= \pm \sqrt[3]{-\frac{1}{x}+C}

\newline

Is Marcus's work correct? If not, what is his mistake?

\newline

1

\newline

(A) Marcus's work is correct.

\newline

2

is incorrect. Marcus didn't integrate

3 y^{2}

\newline

3

is incorrect. Marcus forgot to add a constant to the right-hand side at the end of the process.

\newline

3

is incorrect. The right-hand side of the equation should be

\sqrt[3]{-\frac{1}{x}+C}

Solve the equation.

\newline

\frac{d y}{d x}=-\frac{1}{3} x^{3}

\newline

1

\newline

y=-\frac{x^{4}}{12}+C

\newline

y=-x^{2}+C

\newline

y=-\frac{x^{4}}{12+C}

\newline

y=-x^{2+C}

Solve the equation.

\newline

\frac{d y}{d x}=-\frac{1}{4} e^{x} y^{-2}

\newline

1

\newline

y= \pm \sqrt[3]{-\frac{3 e^{x}}{4}+C}

\newline

y=\sqrt[3]{-\frac{3 e^{x}}{4}+C}

\newline

y= \pm \sqrt[3]{-\frac{3 e^{x}}{4}}+C

\newline

y=\sqrt[3]{-\frac{3 e^{x}}{4}}+C

...