# Intro To Calculus

This article is geared towards the practical application of calculus rather than the theoretical aspect. I won't be deriving the equations that are used here and I won't be showing all the theory behind everything, but I'll give a list of things to look up if you want to know more about where these concepts come from. Also, I've noted that in many other countries the terminology is different from what I'll be using, so don't be outraged if I say derivative instead of differential or something else.

#### Basic Derivatives

This is really the backbone of calculus. Everything else in this article will be based off of this concept. If you want to know where the concept of derivatives comes from then look up "limits" and "tangent lines". Basically, the derivative of a function is the rate of change of one variable with relation to another (usually change in y with relation to x). This is notated in a few ways; the two most common are: f'(x), which is the derivative of f(x) and dy/dx, which notates the derivative of a function as well ('d' representing the Greek letter delta which is often used to represent "change". dy/dx is the change in y with respect to the change in x).

The definition of the slope of a two dimensional function, y=f(x), is the change in y with respect to x (often said as rise over run). Up until calculus you could only approximate the slope of nonlinear functions at certain points. The following is a very basic formula for taking a derivative of a polynomial.

Given that f(x)=ax^{b},

The derivative is f'(x)=abx^{(b-1)}.

A very simple example for this is f(x)=x^{2}. The derivative is f'(x)=2x. Also, you should note that the derivative of an integer (i.e. non-variable) is zero (i.e. f(x) = 2, f'(x) = 0). This is because a constant such as two is actually 2x^{0} (since anything to the power of zero equals one), and the derivative of that is 0*2x^{-1}. And when dealing with an equation of this form (i.e. f( x ) = ax^{b}+cx^d+...), you take the derivative of each term (e.g. f(x)=x^{2} + x^{3}; f'(x)=2x+3x^{2}).

Perhaps you who haven't learned this yet are skeptical about the idea that you can find a slope of a curve this way. Let's use a simple example that you can confirm with your current knowledge, a linear equation.

Let's say we have the equation y=2x+2. That equation is in the form y=mx+b where m is the slope and b is the y-intercept, therefore the slope equals two. Now let's compare to the derivative: y=2x+2; y'=2x^{0}+0*2x^{-1} = 2.

Derivatives can be used for many different practices including geometry, physics, statistics, accounting, and many more. They can even be applied to everyday life. With what we have just learned, we can solve a simple but useful practical application problem. Let's say you have a plot of land bordering a river, and you want to put a fence around it (you don't need to put fencing along the river side). You have enough material to lay 1000 meters of fence. What should the dimensions be if you want to enclose the maximum area with the materials you have?

Here's how you can use calculus to solve it:

You'll end up making two equations (one for perimeter and one for area). First, we need to work with the perimeter equation.

X+2y=1000 (We only need to cover three sides since the river covers the fourth)

You then need to solve that equation for y.

Y=500 - x/2

Now we make our area equation.

A=x*y

We need the problem to be written with only one independent variable to solve it, so take the result from our perimeter equation and substitute y for 500-x/2.

A=x(500-x/2)=500x - (x^{2})/2

In order to find the maximum area, we need to find the maximum point on the area curve. In this instance, that will be when the slope of the curve (i.e. the derivative) is equal to zero. Since this is a quadratic equation there will be exactly one maximum or minimum (maximum in this case since the term with the exponent is negative).

dA/dx=500 - x

Now we set dA/dx to zero and we have: 0=500-x

And therefore x=500. Now just plug that value back into the original perimeter equation and solve for y:

500+2y=1000; 2y=500; y=250

#### More Complex Derivatives

The derivatives we've worked with so far have been fairly easy to calculate, but what do you do if you have a function like this:

f(x) = (3x+1)(4x-1)

You could easily expand that into a nice polynomial, but what if your function is much more complex and you can't? There are a few simple procedures you can use to take derivatives of these types of functions: Product Rule, Quotient Rule, and the Chain Rule.

**Product Rule:**

Given an equation of the form y=f(x)*g(x), you can find the derivative of y using the following formula.

y' = f(x)*g'(x) + f'(x)*g(x)

So let's use our example from above, f(x) = (3x+1)(4x-1)

f'(x) = (3x+1)*(4) + (3)*(4x-1) = (12x+4)+(12x-3) = 24x+1

Let's expand the original f(x) and take the derivative so you can all see that it comes to the same answer. f(x) = (3x+1)(4x-1) = 12x^{2} + x -1

f'(x)= 24x+1

**Quotient Rule:**

Now let's say our equation is f(x) = (3x+1)/(4x-1). We need to use the quotient rule to solve this.

Given an equation of the form y=f(x)/g(x), you can find the derivative of y using the following formula.

y' = ( f'(x)*g(x) - f(x)*g'(x) ) / (g(x))^{2}

In words: The denominator multiplied by the derivative of the numerator minus the numerator multiplied by the derivative of the denominator, all divided by the denominator squared.

Using our example from above, f(x) = (3x+1)/(4x-1), let's apply the quotient rule.

f'(x) = ( (3)*(4x-1) - (3x+1)(4) ) / (4x-1)^{2}

= ((12x-3) - (12x+4)) / (16x^2-8x+1)

=(-7)/(16x^2 - 8x + 1)

**Chain Rule:**

Now, let's say you have the equation y = (4x-1)^{2}. We could expand that and take the derivative of the polynomial, but what if we have an exponent much larger than two? The most effective way to solve the problem is the chain rule. Here is the general formula for the chain rule.

If y = f( g(x) ), then y' is given by:

y' = g'(x) * f'( g(x) )

Let's try applying that to y = (4x-1)^{2}.

The easiest way to go about this now is to replace g(x) with a variable so it looks less confusing. So let's write it as: y = u^{2} where u=4x-1.

y' = u'*2u

Now, replace u with (4x-1) and u' with (4) and you have your answer.

y' = (4)*2(4x-1) = 32x-8

We can check this answer since we know that (4x-1)^{2} = 16x^{2}-8x+1, and the derivative of that is 32x-8.

**Taylor Series Example:**

Here's another example to show how calculus can be useful involving Taylor Series. For this example I'll introduce another derivative formula. The derivative of the natural logarithm is:

f(x)=C*ln(g(x)) where C is some constant and g(x) is a function of x.

f'(x)=(C*g'(x)) / g(x)

So, the derivative of y=ln(x) is y'=1/x.

The Taylor Series is a way to represent complex functions as an infinite series of polynomials. This can be useful in numerical computing when you need to perform many operations on a very ugly equation. The Taylor Series also becomes useful for approximating integrals in higher level calculus. Here is the formula for finding the Taylor approximation of a function:

For those of you already familiar with the Taylor polynomial, note that this particular form is called the Maclaurian Series which is centered about zero.

Also, note that f''(x) is the second derivative of f(x) which is just the derivative of f'(x).

Now let's say you have an ugly function like f(x)= ln(x^{2}+1)+4x+2 and you want to turn it into a polynomial. Let's find the Taylor approximation with two terms (note that using more terms yields a function that is a more accurate approximation).

To solve we need to find all of the pieces needed for the polynomial:

We can easily plug in zero for x and find that f(0)=2.

Now to find f'(0):

f(x)=ln(u)+4x+1

u=x^{2}+1

u'=2x

f'(x)=2x/(x^{2}+1)+4

f'(0)=4

So our new polynomial is (note that it's not a terribly accurate approximation because we only used two terms):

f(x)=2+4x

For those of you interested in learning more about derivatives, you can easily find the formulas online for differentiating other functions such as logarithms, trigonometric functions, hyperbolic functions, etc. If you readers find this article interesting and want to see more, let me know and I can cover the topic of calculus further. Calculus is a vast subject and the basic derivative is only the tip of the ice berg. The next major topic in calculus is integrals (related terms: primitive function, anti-derivative, definite/indefinite integral).

## Comments:

dnatrixene135