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Calculus Help
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
- Positive 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.
The problem presents the typical case of a quotient of two
quantities that go to infinity. In fact,
and
both approach to infinity as
. Which means that
the quotient
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
as
because
,
and
as
.
This trick is applied in general for this type of expressions.
Problem 2. Find
and
if
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:
where
. The 0 on the right hand side
comes from differentiating the constant
. We multiply
the equation by
and we get
which means that
Let's get
. From the last equation we obtain easily that
and now we differentiate this again with respect to
to get:
But
and
. We know put all the
elements together:
and now we simplify the right hand side of the last equation
But by definition:
, so we replace this
into the previous equation:
which gives the final answer.
Check more examples here.
Calculus handout: Here we present you with a free calculus handout that includes calculus and pre-calculus
topics. You can download it here