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Calc II - Midterm 1 Review

Total questions: 13

Worksheet time: 22mins

Name
Class
Date
1.

Find f⁻¹(3) where f(x) = x³ + 2x

a)

0

b)

3

c)

1

d)

-1

2.

Find (f⁻¹)'(3) where f(x) = x³ + 2x

a)

5

b)

1/5

c)

-5

d)

-1/5

3.

Find the inverse function f(x)=(x1)(x+1)f(x)=\frac{(x-1)}{(x+1)}

a)

f1(x)=(x+1)(x1)f^{-1}(x)=\frac{(x+1)}{(x-1)}

b)

f1(x)=(1x)(1+x)f^{-1}(x)=\frac{(1-x)}{(1+x)}

c)

f1(x)= (x+1)(1x)f^{-1}\left(x\right)=\frac{\ \left(x+1\right)}{\left(1-x\right)}

d)

f1(x)= (x+1)(1+x)f^{-1}\left(x\right)=\frac{\ \left(x+1\right)}{\left(1+x\right)}

4.

Use log differentiation to find the derivative of y=(x1)15(x2+1)12y=\frac{(x-1)^{\frac{1}{5}}}{(x^2+1)^{\frac{1}{2}}}

a)

y=(x1)45(x2+1)12x(x1)15(x2+1)32y'=\frac{(x-1)^{-\frac{4}{5}}}{(x^2+1)^{\frac{1}{2}}}-\frac{x(x-1)^{\frac{1}{5}}}{(x^2+1)^{\frac{3}{2}}}

b)

y=15(x1)15(x2+1)12x(x1)15(x2+1)32y'=\frac{1}{5}\frac{(x-1)^{\frac{1}{5}}}{(x^2+1)^{\frac{1}{2}}}-\frac{x(x-1)^{\frac{1}{5}}}{(x^2+1)^{\frac{3}{2}}}

c)

y=15(x1)45(x2+1)12x(x1)15(x2+1)32y'=\frac{1}{5}\frac{(x-1)^{-\frac{4}{5}}}{(x^2+1)^{\frac{1}{2}}}-\frac{x(x-1)^{\frac{1}{5}}}{(x^2+1)^{\frac{3}{2}}}

d)

y=15(x1)45(x2+1)12x(x1)15(x2+1)12y'=\frac{1}{5}\frac{(x-1)^{-\frac{4}{5}}}{(x^2+1)^{\frac{1}{2}}}-\frac{x(x-1)^{\frac{1}{5}}}{(x^2+1)^{\frac{1}{2}}}

5.

Differentiate with respect to x:

xπe(x2+1)x^{\pi}e^{(x^2+1)}

a)

πx(π1)e(x2+1)2x1+πe(x2+1)\pi x^{(\pi-1)}e^{(x^2+1)}-2x^{1+\pi}e^{(x^2+1)}

b)

πx(π1)e(x2+1)+2x1+πe(x2+1)\pi x^{(\pi-1)}e^{(x^2+1)}+2x^{1+\pi}e^{(x^2+1)}

c)

πx(π1)e(x2+1)\pi x^{(\pi-1)}e^{(x^2+1)}

d)

2x1+πe(x2+1)2x^{1+\pi}e^{(x^2+1)}

6.

Differentiate with respect to x:

ln(x2+3)\ln(x^2+3)

a)

x(x2+3)\frac{x}{(x^2+3)}

b)

1(x2+3)\frac{1}{\left(x^2+3\right)}

c)

2(x2+3)\frac{2}{(x^2+3)}

d)

2x(x2+3)\frac{2x}{(x^2+3)}

7.

Differentiate with respect to x:

(ln(x3+1))ex\left(\ln(x^3+1)\right)^{e^x}

a)

eexln(ln(x3+1))ln(x3+1) 3x2x3+1ex \frac{e^{e^x\ln\left(\ln\left(x^3+1\right)\right)}}{\ln\left(x^3+1\right)}\ \frac{3x^2}{x^3+1}e^{x\ } +ln(ln(x3+1))exeexln(ln(x3+1))+\ln\left(\ln\left(x^3+1\right)\right)e^xe^{e^x\ln\left(\ln\left(x^3+1\right)\right)}

b)

ex(ln(x3+1))ex1(x3+1)3x2\frac{e^x\left(\ln\left(x^3+1\right)\right)^{e^x-1}}{\left(x^3+1\right)}3x^2

c)

eln(x3+1)ex3x2exx3+1\frac{e^{\ln\left(x^3+1\right)e^x}3x^2e^x}{x^3+1}

d)

eln(x3+1)ex 3x2(x3+1)ex \frac{e^{\ln\left(x^3+1\right)e^x}\ 3x^2}{\left(x^3+1\right)}e^{x\ }

+eln(x3+1)exln(x3+1)ex+e^{\ln\left(x^3+1\right)e^x}\ln\left(x^3+1\right)e^x

8.

 cos(x)esin(x)+1 dx\int\ \cos\left(x\right)e^{\sin\left(x\right)+1}\ dx

a)

eu+Ce^u+C

b)

esin(x)+Ce^{\sin\left(x\right)}+C

c)

esin(x)+1+Ce^{\sin\left(x\right)+1}+C

d)

esin(x)+Ce^{\sin\left(x\right)+C}

9.

Integrate: ex(ex+1)dx\int\frac{e^x}{(e^x+1)}dx

a)

ex(ex+1)+C\frac{e^x}{(e^x+1)}+C

b)

ln(ex+1)+C\ln(e^x+1)+C

c)

lnex1+C\ln\left|e^x-1\right|+C

d)

ex+Ce^x+C

10.

Integrate:  1425x2 dx\int\ \frac{1}{\sqrt{4-25x^2}}\ dx

a)

 15sin1(5x2) +C\frac{\ 1}{5}\sin^{-1}\left(\frac{5x}{2}\right)\ +C

b)

 15tan1(5x2) +C\frac{\ 1}{5}\tan^{-1}\left(\frac{5x}{2}\right)\ +C

c)


 15sec1(5x2) +C\frac{\ 1}{5}\sec^{-1}\left(\frac{5x}{2}\right)\ +C

d)

 12sin1(2x5) +C\frac{\ 1}{2}\sin^{-1}\left(\frac{2x}{5}\right)\ +C

11.

John deposited $500 in a bank account. How much money will John have after 2 years with 5% interest compounded continuously?

a)

500e(0.05)2500e^{\left(0.05\right)2} dollars

b)

500e52500e^{5\cdot2} dollars

c)

500(0.05)2500\left(0.05\right)2 dollars

d)

0.05e50020.05e^{500\cdot2} dollars

12.

Which of the following is an integrating factor for the differential equation

t dydt+2y=t2t+1t\frac{\ dy}{dt}+2y=t^2-t+1

a)

e2tdte^{\int\frac{2}{t}dt}

b)

e2dte^{\int2dt}

c)

e2tdt e^{\int2tdt\ }

d)

etdte^{\int tdt}

13.

Use the general solution

y(t)=14t213t+12+ct2y(t)=\frac{1}{4}t^2−\frac{1}{3}t+\frac{1}{2}+\frac{c}{t^2}

of the previous differential equation to solve the initial value problem y(1)=12y\left(1\right)=\frac{1}{2} .

a)

y(t)=14t213t+12+7t2y(t)=\frac{1}{4}t^2−\frac{1}{3}t+\frac{1}{2}+\frac{7}{t^2}

b)

y(t)=14t213t+12+112t2y(t)=\frac{1}{4}t^2−\frac{1}{3}t+\frac{1}{2}+\frac{1}{12t^2}

c)

y(t)=14t213t+12+12t2y(t)=\frac{1}{4}t^2−\frac{1}{3}t+\frac{1}{2}+\frac{12}{t^2}

d)

y(t)=14t213t+12+17t2y(t)=\frac{1}{4}t^2−\frac{1}{3}t+\frac{1}{2}+\frac{1}{7t^2}