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Calculus is the mathematics of all things that change. It revolves around utilizing derivatives and integrals to solve problems. To understand derivatives and integrals, consider a grid with an origin at point (0, 0). Moving one space to the left yields position (-1, 0) followed by moving two spaces up to be at space (-1, 2). Position is considered the original equation and will often show the y value as dependent on the x value. This example does not have enough information to determine y at any point as the slope has not been defined. For this example, the slope will be 3. That yields y = 3x + 5. What does that mean? The position of y begins 5 paces above the origin and will increase three paces for every increase in x. The derivative of this equation will be the rate of change, or the slope. Most equations are not this straightforward.
Consider the equation y = 3x^2 + 7. Because 7 is a constant value, it has no bearing on finding the derivative. Starting at x = 0, the first five values of y would be 0, 10, 19, 34, and 55. Because the change is not constant, a derivative must be found. To find the derivative, students would use one of two methods. The first method is the definition of a derivative.
This example illustrates how the constant 7 cancels itself out and h can be divided out. The result is a first order derivate in the form y’= 6x. Other ways to write this equation are y’ = y1 = dy/dx or to refer to this as velocity v or v(x). The derivative obtained can be used to determine the rate of change at any value of x. For example, when x is 0, the position is not changing at all. When x is 1, the position is changing at a rate of 6 paces upward. When x is 2, the position is changing at a rate of 12 paces upward. This method can be repeated indefinitely until the derivative is 0.
The second method is to use “rules” or “tricks” to determine the derivative. When finding the derivative, operations are maintained. That means if two values are being multiplied or added, the derivative will reflect that. The first “trick” is all lone constants have derivatives of 0. This was illustrated in the first method when the 7 cancelled itself out. The second “trick” was observed when finding the derivative of y = 3x + 5: the coefficient in front of x will remain as a constant. A parabola is more complicated because x has an exponent. That is the third “trick:” the derivative of an exponent xa will always be axa-1. In this example, the derivative of x2 will be 2x. Maintaining the operations, y’ = 3*(2x) = 0, or y’ = 6x.
The first method is less memorization than the second method, but it becomes time consuming with extra work. This is undesirable during tests, especially the AP or IB exam when every second counts. The second method requires more practice to become familiar with every set of “rules” and “tricks,” but is easier to do in the long run with complicated equations.
Integration is the opposite of a derivative and does not have a standard definition like method one for derivation. Therefore, integration is usually learned after students are proficient in finding and utilizing derivatives. Another reason is because all the “tricks” and “rules” of integration are the opposite of derivation. Using the example y = 3×2 + 7, when a constant like 7 is integrated, it becomes a coefficient like 7x. 3×2 will be split in two parts. Remember the rule for deriving an exponent xa is it becomes axa-1 so the integral of x2 becomes x3. When deriving x3, the 3 would be dropped in front of the x. This appears in our equation 3×2 + 7 which means it does not need to be counteracted. However, if our equation was 5×2, there would need to be a coefficient of 1/3 placed before x in the form of (5/3)x3 as the derivative of that would be (5/3)*3×2. Maintaining operations and bringing the integral together, the solution takes the form:
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