Understanding Derivatives

Understanding Derivatives Patrick Braga May 17, 2009 Consider the graph of a linear equation. Being that it is simply a straight line, it has a consistent slope throughout its entirity. Now consider the graph of a quadratic equation. Its slope changes continuously: at points, it may be steeper or less steep. The slope of any given point of the plotted equation is equal to the slope of a line tangent to the curve at that exact point. Let it be considered now that a point (a, f (a)) is selected on the curve of the graphed quadratic function and that a nearby point h units away is also (a) selected: (a + h, f (a + h)). The slope f (a+h)−f represents the secant line (a+h)−a defined between the two said points. It should be agreed upon that when h = 0, the slope of the curve at the point (a, f (a)) will have been calculated. Knowing this, the best way to approximate the slope of the curve at the moment is by finding its limit upon approaching 0. lim

x→0

f (a + h) − f (a) (a + h) − a

When the slope of the secant line approaches that of the tangent line, revealing the slope of the curve on the point (a, f (a)). In fact, that is the definition of the derivative: the slope of a curve at an exact curve. Let’s apply this concept. Given the function f (x) =2 , what is the slope of the curve when x = 3? When 3 is inputted to the function, the output is 9. This gives us the coordinate (3, 9). Let’s continue using the idea of a nearby point h units away, giving us the coordinate (3 + h, (3 + h)2 ). This results in the following slope: (3+h)2 −9 (3+h)−3

This is the simplified slope: 6+h Now, we can find the instantaneous slope of the curve at (3, 9) by finding the limit of h as it approaches 0 (that is, the limit of a distance between the two points): lim (6 + h) = 6

h→0

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This however, is the slope specific only to (3, 9). Instead of specifying the slope to this one coordinate in lim (6 + h) = 6, it can be generalized to h→0

f (x + h) − f (x) lim , which can hence be simplified to lim (2x+h), which equals h→0 h→0 h 2x. This derivative of f (x) should always be denoted as f 0 (x). The derivate function of f (x) = x2 is, then, f 0 (x) = 2x. While the limit formula can be used to find the derivative of a function with more than one variable, a property known as the “product rule” can also be applied. Let’s take a more complex polynomial function: f (x) = 2x2 + 4x + 8 We want to find its derivative: ∂2x2 +4x ∂x

To apply the product rule, the first thing that needs to be done is to separate the polynomial into various separate monomials: 2x2 and 4x. In each of these monomials, the coefficient is multiplied by the value of the exponent and 1 is subtracted from the exponent’s value, resulting in the following monomials: 4x and 4. From here, the derivative function can be made: f 0 (x) = 4x + 4 Or, in an alternate simplified form: f 0 (x) = 4(x + 1) Let’s check to make sure that the derivative function was calculated correctly, following the previously stated method. Here is the slope of f (x) = 2x2 + 4x + 8 between x and x + h: m=

f (x+h)−f (x) (x+h)−x

=

f (x+h)−f (x) h

=

(2[x+h]2 +4[x+h])−(2[x]2 +4[x]) h

= 2h + 4x + 4

Now we can solve for the derivative function by seeing the limit of the slope as it approaches zero. f 0 (x) = lim 2h + 4x + 4 −→ f 0 (x) = 4x + 4 h→0

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