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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 problems to help you determine if you are ready.
If you struggle with these problems, don't worry. Use 17calculus to learn what you need to fill in the gaps. Or use these problems to determine where you need to go to refresh your calculus 1 skills, if it's been a while since you took calculus 1.

Here are the main topics you need for calculus 2.
limit techniques 
1. basic limits 
2. finite limits 
3. infinite limits 
differentiation techniques 
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 
integration techniques 
1. basic integration 
2. integration of basic trig functions 
3. integration by substitution 
applied integration 
4. area between curves 
It also helps to go through implicit differentiation and logarithmic differentiation to practice the product rule and the chain rule. 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.
Okay, let's get started with some practice problems.
For explanation of this technique and more practice problems, go to the 17calculus implicit differentiation page.
Instructions  Unless otherwise instructed, calculate \(dy/dx\) by implicit differentiation. Give your answers in simplified, completely factored form.
Question 1 
\(x^2+3xy+y^3=10\)



\(\displaystyle{ \frac{dy}{dx} = \frac{1}{3}\left[\frac{2x+3y}{x+y^2}\right] }\)

First, take the derivative of both sides with respect to x. 
\(\displaystyle{\frac{d}{dx}\left[x^2+3xy+y^3\right] = \frac{d}{dx}[10]}\) 
Distribute the derivative. 
\(\displaystyle{\frac{d}{dx}\left[x^2\right] + \frac{d}{dx}\left[3xy\right] + \frac{d}{dx}\left[y^3\right] = \frac{d}{dx}[10]}\) 
Apply the derivative to each term. The term \(d[3xy]/dx\) requires the use of the product rule. 
\(\displaystyle{2x + \frac{3d[xy]}{dx} + 3y^2\frac{dy}{dx} = 0 }\) 
\(\displaystyle{ 2x + 3 \left[ x\frac{dy}{dx} +y(1)\right] +3y^2\frac{dy}{dx} = 0 }\) 
Separate the bracketed term. 
\(\displaystyle{ 2x + 3x\frac{dy}{dx} + 3y +3y^2\frac{dy}{dx} = 0 }\) 
Keep the terms containing \(dy/dx\) on the left of the equal sign. Move the other terms to the right. 
\(\displaystyle{ 3x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 2x3y }\) 
Factor \(dy/dx\) from all terms on the left. 
\(\displaystyle{ (3x+3y^2)\frac{dy}{dx} = 2x3y }\) 
Divide both sides by the \(3x+3y^2\) term to solve for \(dy/dx\). 
\(\displaystyle{\frac{dy}{dx} = \frac{2x3y}{3x+3y^2}}\) 
Factor out \(1/3\) to simplify. This gives us our final answer. 
\(\displaystyle{ \frac{dy}{dx} = \frac{1}{3}\left[\frac{2x+3y}{x+y^2}\right] }\) 
Question 1 Final Answer 
\(\displaystyle{ \frac{dy}{dx} = \frac{1}{3}\left[\frac{2x+3y}{x+y^2}\right] }\)

Question 2 
\( y^2=(xy)(x^2+y) \)



\(\displaystyle{ \frac{dy}{dx} = \frac{3x^22xy+y}{x^2x+4y} }\)

Take the derivative of both sides with respect to x. 
\(\displaystyle{ \frac{d}{dx}[y^2] = \frac{d}{dx}[(xy)(x^2+y)] }\) 
We need to use the product rule on the term on the right. Note: If desired, you can multiply out this term before taking the derivative. In this particular problem, it doesn't matter. However, in general, you will probably want to keep the terms factored since it may make it easier later in the problem to factor. 
\(\displaystyle{ 2y \frac{dy}{dx} = (xy)\left[ 2x+\frac{dy}{dx}\right] + (x^2+y)\left[ 1\frac{dy}{dx} \right] }\) 
Distribute terms. 
\(\displaystyle{ 2y \frac{dy}{dx} = 2x(xy) + (xy)\frac{dy}{dx} + x^2+y  (x^2+y)\frac{dy}{dx} }\) 
Move terms with \(dy/dx\) to the left and all other terms to the right. 
\(\displaystyle{ 2y \frac{dy}{dx} (xy)\frac{dy}{dx} +(x^2+y)\frac{dy}{dx} = 2x(xy)+x^2+y }\) 
Factor out the \(dy/dx\). 
\(\displaystyle{ (2yx+y+x^2+y)\frac{dy}{dx} = 2x^22xy+x^2+y }\) 
Solve for \(dy/dx\) and simplify. 
\(\displaystyle{ \frac{dy}{dx} = \frac{3x^22xy+y}{x^2x+4y} }\) 
Question 2 Final Answer 
\(\displaystyle{ \frac{dy}{dx} = \frac{3x^22xy+y}{x^2x+4y} }\)

Question 3 
\(\cos x^2 = xe^y\)



\(\displaystyle{ \frac{dy}{dx} = \frac{e^y + 2x\sin x^2}{xe^y} }\)

Take the derivative of both sides with respect to x. 
\(\displaystyle{\frac{d}{dx}[\cos x^2] = \frac{d}{dx}[xe^y] }\) 
The left side requires the chain rule and the right side requires the product rule. 
\(\displaystyle{\sin x^2(2x) = xe^y \frac{dy}{dx} + e^y(1)}\) 
Move the terms with \(dy/dx\) to the left and all other terms to the right. 
\(\displaystyle{ xe^y \frac{dy}{dx} = e^y2x\sin x^2 }\) 
Solve for \(dy/dx\). 
\(\displaystyle{ \frac{dy}{dx} = \frac{e^y2x\sin x^2}{xe^y} }\) 
Question 3 Final Answer 
\(\displaystyle{ \frac{dy}{dx} = \frac{e^y + 2x\sin x^2}{xe^y} }\)

Question 4 
\( ye^x+xe^y = xy \)



\(\displaystyle{ \frac{dy}{dx} = \frac{yye^xe^y}{e^x+xe^yx} }\)

Take the derivative of both sides with respect to x. 
\(\displaystyle{ \frac{d}{dx}[ye^x+xe^y] = \frac{d}{dx}[xy] }\) 
Take the derivative of each term. All derivaties require the product rule. 
\(\displaystyle{ ye^x + e^x \frac{dy}{dx} + xe^y \frac{dy}{dx} + e^y (1) = }\) \(\displaystyle{ x \frac{dy}{dx} +y(1) }\) 
Move the terms with \(dy/dx\) to the left and all other terms to the right. 
\(\displaystyle{ e^x \frac{dy}{dx} + xe^y \frac{dy}{dx} x \frac{dy}{dx} = yye^xe^y }\) 
Factor out the \(dy/dx\) from all terms on the left. 
\(\displaystyle{ [e^x+xe^yx]\frac{dy}{dx} = yye^xe^y }\) 
Solve for \(dy/dx\). 
\(\displaystyle{ \frac{dy}{dx} = \frac{yye^xe^y}{e^x+xe^yx} }\) 
Question 4 Final Answer 
\(\displaystyle{ \frac{dy}{dx} = \frac{yye^xe^y}{e^x+xe^yx} }\)

Question 5 
\(\sqrt{xy}=x4y\)



\(\displaystyle{\frac{dy}{dx} = \frac{2\sqrt{xy}y}{x+8\sqrt{xy}} }\)

First, write the square root as a power. This will allow us to take the derivative more easily. 
\((xy)^{1/2}=x4y\) 
Take the derivative of both sides of the equation with respect to x. 
\(\displaystyle{ \frac{d}{dx}\left[(xy)^{1/2}\right] = \frac{d}{dx}[x4y]}\) 
Use the product rule on the left side and take the derivative of each term on the right. 
\(\displaystyle{\frac{1}{2}(xy)^{1/2}\left[x \frac{dy}{dx}+y(1)\right] = 14\frac{dy}{dx}}\) 
Distribute the term on the left and multiply both sides by 2. 
\(\displaystyle{\frac{x}{\sqrt{xy}}\frac{dy}{dx} + \frac{y}{\sqrt{xy}} = 2  8 \frac{dy}{dx} }\) 
Rearrange the terms so that the terms with \(dy/dx\) are on the left side of the equal sign and the other terms are on the right. 
\(\displaystyle{\frac{x}{\sqrt{xy}} \frac{dy}{dx} + 8 \frac{dy}{dx} = 2  \frac{y}{\sqrt{xy}} }\) 
Factor out the \(dy/dx\) from all terms on the left and get common denominators for all terms. 
\(\displaystyle{\frac{dy}{dx}\left[ \frac{x}{\sqrt{xy}} +8 \frac{\sqrt{xy}}{\sqrt{xy}} \right] = \frac{2\sqrt{xy}}{\sqrt{xy}}  \frac{y}{\sqrt{xy}}}\) 
Divide both sides by the term on the left and simplify. 
\(\displaystyle{\frac{dy}{dx} = \left[ \frac{2\sqrt{xy}y}{\sqrt{xy}} \right] \left[ \frac{\sqrt{xy}}{x+8\sqrt{xy}} \right] = }\) \(\displaystyle{ \frac{2\sqrt{xy}y}{x+8\sqrt{xy}} }\) 
Note: Although this is fairly simplified, some instructors may ask you to factor out a \(\sqrt{y}/\sqrt{x}\). 
\(\displaystyle{\frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}}\left[ \frac{2\sqrt{x}\sqrt{y}}{\sqrt{x}+8\sqrt{y}} \right]}\) 
This last form is completely factored but not as simplified as the previous form (in our opinion). Ask your instructor which form they prefer. We have listed the previous form as the final answer. 
Question 5 Final Answer 
\(\displaystyle{\frac{dy}{dx} = \frac{2\sqrt{xy}y}{x+8\sqrt{xy}} }\)
