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This is a list of subjects you may be interested on

Functions
- Polynomials
- Rational Functions
- Logarithms
- Exponentials
- Trigometric

Limits
- Direct Substitution Property
- Undetermined Forms
- Trigometric Limits
- Factorization
- Side Limits
- L'Hopital Rule

Continuity
- Connection with limits
- Algebraic Properties
- Intermediate Value Theorem

Derivatives
- Tangent Line
- Product Rule
- Quotient Rule
- Chain Rule
- Related Rates
- Mean Value Theorem
- Maximum and Minimum values

Integrals
- Antiderivatives
- Fundamental Theorem of Calculus
- Substitutions and change of variables
- Partial Fractions
- Improper Integrals

Series
- Postive Series
- Convergence criteria
- Alternating Series

We know how it feels to deal with hard calculus problems. With our help, you'll get that extra edge you need! No need to sweat with poorly explained problems from the books. We can provide solutions that suit your own needs. See some examples:

Problem 1.Find

Solution.

$\displaystyle \lim_{n\rightarrow \infty} \frac{2n^2+1}{3n^2-5n+2}$

The problem presents the typical case of a quotient of two quantities that go to infinity. In fact, and both approach to infinity as $n\rightarrow\infty$ . Which means that the quotient

$\displaystyle \frac{2n^2+1}{3n^2-5n+2}$

is an undetermined form. Somehow we have to cancel the "bad" part, if possible. Notice that since both and are big, there's a chance for cancelation, so the quotient could be finite.

The trick in this case is to divide by the highest power on the expression (from both numerator and denominator). In this case we will divide both numerator and denominator by . We get

$\displaystyle \frac{2n^2+1}{3n^2-5n+2}= \frac{2+\frac{1}{n^2}}{ 3-\frac{5}{n}+\frac{2}{n^2} }\rightarrow \frac{2}{3}$

as $n\rightarrow\infty$ because $\frac{1}{n^2}\rightarrow 0$ , $\frac{5}{n}\rightarrow 0$ and $\frac{2}{n^2} \rightarrow 0$ as $n\rightarrow\infty$ .

This trick is applied in general for this type of expressions. $\Box$

Problem 2. Find $\frac{dy}{dx}$ and $\frac{d^2 y}{dx^2}$ if

$\displaystyle x^{2/3}+y^{2/3}=a^{2/3}, \qquad a>0$

Solution. Let's differentiate both sides of the equality. Let's recall that we are assuming that . That means that the equation gives us a way to solve as a function of , but we still don't know how to express that function. That assumption requires further justification (meaning, it's not always granted), but we'll take it for granted. So, since we get using the chain rule:

$\displaystyle \frac{2}{3} x^{-1/2}+\frac{2}{3} y^{-1/3} y'=0$

where $y'\equiv \frac{dy}{dx}$ . The 0 on the right hand side comes from differentiating the constant $a^{2/3}$ . We multiply the equation by and we get

$\displaystyle x^{-1/2}+ y^{-1/3} y'=0$

which means that

$\displaystyle \frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3}$

Let's get . From the last equation we obtain easily that

$\displaystyle x\left(\frac{dy}{dx}\right)^3=-y$

and now we differentiate this again with respect to to get:

$\displaystyle \left(\frac{dy}{dx}\right)^3+3x\left(\frac{dy}{dx}\right)^2 y''=-y'$

But $\left(\frac{dy}{dx}\right)^3=-y/x$ and $3x\left(\frac{dy}{dx}\right)^2=3x(y/x)^{2/3}$ . We know put all the elements together:

$\displaystyle y''=\frac{y/x+(y/x)^{1/3}}{3x(y/x)^{2/3}}$

and now we simplify the right hand side of the last equation

$\displaystyle y''=\frac{(y/x)^{2/3}+1}{3x(y/x)^{1/3}}=\frac{y^{2/3}+x^{2/3}}{3x^{4/3}y^{1/3}}$

But by definition: $x^{2/3}+y^{2/3}=a^{2/3}$ , so we replace this into the previous equation:

$\displaystyle y''=\frac{a^{2/3}}{3x^{4/3}y^{1/3}}$

which gives the final answer. $\Box$

Check more examples here.

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