## 17Calculus - Are You Ready For Calculus 2?

So, you finished calculus 1. Congratulations! Do you think you are ready for calculus 2? Let's find out. Here are some practice problems from calculus 1 that use techniques that you need for calculus 2. Calculus 2 is the hardest of the three calculus courses and your calculus 1 skills need to be sharp. So here are some steps to help you determine if you are ready.

Here are the main topics you need for calculus 2.

limits

1. basic limits

2. finite limits

3. infinite limits

derivatives

1. basic power and trig rules

2. product rule

3. quotient rule

4. chain rule

applied differentiation

5. maxima and minima

6. equations of tangent lines

integrals

1. basic integration

2. integration of basic trig functions

3. integration by substitution

applied integration

4. area between curves

The chain rule is the single most important and most used rule of all of the derivative rules.
Integration by substitution is the single most important and most used rule of all the integration rules.

Here are the recommended steps to go through to make sure you are prepared for calculus 2.

Algebra

1. If you struggled with algebra in calculus 1, which many students do, go to the precalculus section to review.

Limits

2. You will be doing infinite limits in calculus 2, as well as a few finite limits. You also need a good understanding of basic limits to work these problems.

Derivatives

3. Make sure you are strong with the product rule, quotient rule and the chain rule. We don't mean just know what they are. We mean be able to do them in your sleep, almost without thinking and without looking them up in the textbook.

4. Do you remember the derivatives of exponential functions, logarithms and all the trig functions? If not, review each page and work some practice problems.

5. Once you are comfortable with the previous two steps, work some logarithmic differentiation problems. These will give you plenty of practice working with logarithms, which is the one topic almost all calculus students struggle with.

6. You need to be able to find the equation of a tangent line and use the derivative to find extrema.

Integrals

7. You need to know and really understand how to use integration by substitution. It is the one technique you will use in almost every integration problem you work. You need to be able to use this technique on both indefinite and definite integrals using correct notation.

8. You will be doing a lot of integration in calculus 2, so make sure you know how to integrate functions with exponentials, logarithms and basic trig functions.

9. It is important to know how to find the area under a curve since calculus 2 builds on that concept to find area between functions.

Okay, so if you have gone through all these steps and you feel confident, then you are well on your way to start calculus 2.
As usual, you need to check with your instructor. They may expect you to know more or different material than is listed here. But these steps should get you started.

You CAN Ace Calculus

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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