Posts Tagged ‘differentiation’
Thursday, September 10th, 2009
How to Solve an Integration Problem by Trigonometric Substitution
Description
This video clearly illustrates how to solve an integration problem using trigonometric substitution. One example problem is provided in the video.
Overview
Trigonometric substitution works the same way as normal substitution, only you substitute in trigonometric functions, and each trigonometric function can only be substituted for a particular pattern. These are the patterns to watch for and what you can substitute in for them:
a^2 – x^2 uses x = a * sin(theta)
a^2 + x^2 uses x = a * tan(theta)
x^2 – a^2 uses x = a * sec(theta)
All of these also have the option of including a square root with them, but it doesn’t matter – you can use the substitution without the square root. Normally after finding the x value you will take a derivative so you have the value of dx To find other values needed,SOHCAHTOA is often used with a right triangle picture. Once all of your values have been solved for, you can plug them all back into your original equation and solve.
Tags: antiderivatives, antidifferentiation, Calculus, derivatives, differentiation, integrals, integration by substitution, Math, secant, sine, tangent, trig functions, trigonometric substitution, trigonometry
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Thursday, September 10th, 2009
How to Solve an Integration Problem by Substitution
Description
This video shows an example of how to solve the same problem with and without substitution, to express how much easier substitution is. This video provides several example problems with clear explanations and solutions for each one.
Overview
The substitution rule comes in very useful no matter how you are trying to solve an integral. When you are presented with something that looks like you may need the chain rule to solve it, or you are presented with a polynomial term in the denominator of your integral, then you can use u-substitution. It is called that because the variable u is the most common one used. Once you have chosen a value for u, find the derivative and write it as du/dx = n, with n being the number value of du/dx. Now, move the du over so that you are left with a value for dx. Now skip back to your original problem. Once you replace a term with u, you are normally left with a common integral, which is easy to solve. Multiply that by dx, just like in the problem. dx will be expressed as what you just solved for. If this is a multiplication problem, move all numbers to the outside and leave yourself with just the variables. Now you should be able to solve this like a normal integration problem.
Tags: antiderivatives, antidifferentiation, Calculus, derivatives, differentiation, integrals, integration by substitution, Math, u-substitution
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Thursday, September 10th, 2009
How to Solve an Integration Problem by Parts
Description
Step-by-step tutorial of the integration by parts method. Several examples are provided in this video. Knowledge of integration by parts is required in calculus.
Overview
Very often, you will see an integral that is also a multiplication problem. There was such a thing as the product rule for derivatives, but what about for anti-derivatives? There is some sort of reverse product rule, and that is what integration by parts is. Sometimes it is also called u-substitution, but the term is misleading because substitution is something else. A basic integration by parts problem is set up like this:
You are allowed to choose which value is your u and which value is your dv. Choosing involves something called LIATE – Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric, Exponential. Put this on a ladder scale, with L being first and E being last. Each term you are multiplying falls on one of the parts of this chart. The highest one will be your u, and the lower one will be your dv. Sometimes you will have to perform integration by parts more than once in a problem.
Tags: antiderivatives, antidifferentiation, Calculus, derivatives, differentiation, integrals, integration by parts, LIATE, Math
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Thursday, September 10th, 2009
The Definition of an Integral
Description
This video describes what an integral or an anti-derivative is. Several examples are provided in this video. Knowledge of integrals is needed to pass calculus.
Overview
An integral is the opposite of a derivative, and therefore is often called an antiderivative. While with a differentiation problem, you need to find a derivative, if you are solving an integral you are already given a derivative and your goal is to revert it back to it\’s common form.
Example: The derivative of 2x is 2. The antiderivative or integral of 2 would be 2x.
You should already have several derivatives memorized. This makes starting to solve integrals much easier, because if you have the derivative memorixed all you need to ask yourself is \”what is that a derivative of?\” and you have your answer. For reference, the integral symbol looks like a very stretched out S symbol, which is supposed to be an italic I.
Tags: antiderivatives, antidifferentiation, Calculus, derivatives, differentiation, integrals, Math
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Friday, September 4th, 2009
How to Solve Derivatives with Trigonometric Functions
Description
This video shows the basic trigonometric functions and their derivatives. Content is laid out in an organized and easy to follow manner.
Overview
Trigonometric functions, also known as just trig functions, are very common – they are sine, cosine, tangent, secant, cotangent, cosecant. These are derivatives that you should have memorized, because there is no good way to solve for them.
d/dx sin(x) = cos(x)
d/dx cos(x) = -sin(x)
d/dx tan(x) = sec^2(x)
d/dx sec(x) = sec(x) * tan(x)
d/dx csc(x) = -csc(x) * cot(x)
d/dx cot(x) = -csc^2(x)
Tags: Calculus, cosecant, cosine, cotangent, derivative, derivatives, differentiation, Math, secant, sine, tangent, trig, trig functions, trigonometric, trigonometric functions, trigonometry
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Friday, September 4th, 2009
How to Solve Derivatives with Logarithmic Functions
Description
This video covers the basic exponential and logarithmic functions, and then shows several sample problems. Many example problems of solving for natural logs are provided in the video. This video also shows the relationship between exponents and natural logs.
Overview
Derivatives with logarithms are rather easy to solve. The first of these is the regular exponent. The exponent is an exponential function, not a logarithmic function, but the two are related. The solution for an exponent, e^x, is:
d/dx (e^x) = e^x
Remember that the exponential solution only works when the variable x is the exponent, and to not use this when you should be using the power rule. The two look very similar and it is easy to mix them up. A problem using the power rule looks like x^2. A problem using the exponential rule looks like 2^x.
The next of these is a natural log. A natural log is a logarithm in its natural form. It is not necessary to understand natural logs to solve a derivative problem with natural logs. This should be a derivative that you have memorized. The natural log is solved like this:
d/dx [ln(x)] = 1 / x
In other words, to solve a natural log simply put one over whatever is inside the parenthesis. Remember that it may be necessary to use other rules with this rule.
Tags: Calculus, derivative, derivatives, differentiation, exponential, exponents, logarithmic, logarithms, logs, Math, natural logarithms, natural logs
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Friday, September 4th, 2009
How to Solve Derivatives Using the Chain Rule
Description
This video provides an example problem of the chain rule used with several other rules of calculus. Steps are organized in an easy-to-follow manner.
Overview
The chain rule is a rule in calculus that can be used to solve derivatives. It is not a complicated rule, but can still be very confuding when first learning it. The chain rule states that:
d/dx [f(n)] = f’(n) * n’
In other words, the derivative of the outside of the equation is done, leaving the inside intact. Then the derivative of the inside is multiplied to it. The chain rule is used every time sine, cosine, tangent, cosecant, cotangent, and secant are in a problem, and normally when a logarithm (or natural log) is involved. It is sometimes easier to split a problem into two seperate parts before solving.
Example: sin(2x) –> d/dx [sin(x)] = cos(x), d/dx (2x) = 2
Therefore: d/dx sin(2x) = cos(2x) * 2
Tags: Calculus, chain, chain rule, derivative, derivatives, differentiation, Math, multiplication
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Friday, September 4th, 2009
How to Solve Derivatives Using the Quotient Rule
Description
This video goes over the quotient rule and how to use it. It provides many sample problems and shows how to solve derivatives with the quotient rule in conjunction with other derivative rules.
Overview
The quotient rule is a rule in calculus that can help you solve derivatives that are divided. The basic form of the quotient rule is:
d/dx (u / v) = [(v * u') - (u * v')] / (v^2)
This translates to be the the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all over the denominator squared. To remember the quotient rule, you can also use the phrase “low di-high minus high d-low, square the bottom and away we go”
Tags: Calculus, derivative, derivatives, differentiation, divide, division, Math, quotient, quotient rule
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Friday, September 4th, 2009
How to Solve Derivatives Using the Product Rule
Description
This video shows when to use the product rule, how to use the product rule, and why you should use the product rule. Not only does it give several sample problems but it explains the entire process of how the product rule actually works, and why it helps you find the right answer.
Overview
The product rule is a rule in calculus that allows you to solve derivatives that are multiplied. The basic form of the product rule is:
d/dx (u * v) = (u * v’) + (v * u’)
In other words, the solution is the first term mulitplied by the derivative of the second term, plus the second term multiplied by the derivative of the first term. It is often required to use the product rule in combination with several other rules.
Tags: Calculus, derivative, derivatives, differentiation, Math, multiplication, product, product rule
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Friday, September 4th, 2009
How to Solve Derivatives Using the Power Rule
Description
This video explains both the Power Rule and the Constant Rule in-depth, and illustrates the difference between different functions with power rules on a graph. It provides several example problems that could be solved using the power rule.
Overview
The Power Rule is a rule in calculus that allows you to solve derivatives. The Power Rule deals with exponents, or powers. The simple power rule states that:
d/dx (x^n) = nx^(n – 1)
In other words, the number of the exponent gets placed in front of x, and then the exponent gets subtracted by 1. An interesting thing about the Power Rule is the \”chain of command\”. The power rule will be easier to use if you memorize this:
d/dx (x^0) = 0
d/dx (x^1) = 1
d/dx (x^2) = 2x
d/dx (x^3) = 3x^2
Tags: Calculus, derivative, derivatives, differentiation, Math, power, power rule
Posted in Calculus | No Comments »
