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Calculus is my favorite part of mathematics. Sure, algebra is a close-second (<3 linear equations), but calculus just rips algebra to shreds.

One reason for this is that calculus introduces a quirky new concept:

Derivatives are (taken from Wikipedia):

*"*The derivative of a function *f*(*x*) of a variable *x* is a measure of the rate at which the value of the function changes with respect to the change of the variable."

Most people learn derivatives with the boring "actual definition", but I learned derivatives (and pretty much all of calculus) with just rules.

**d/dx(c)=0**

Now, let me break this down for you:

**d/dx** is the derivative operator. It tells you "Hey! You should take the derivative of this!" (note that it is written as d over dx (d on top of dx with a line between them))

**(c)** tells us that we want to take the derivative of c. In this case, c represents any constant.

**=0****.... do I have to explain this?**

You may be asking, why is it 0?

Well, calculus is the study of change.

Derivatives are the rate of change with respect to a variable.

*c *is a constant, which means it doesn't change.

All numbers (1, 3, -8, 3.14159...) that have a value that doesn't change are constants.

Since the derivative is the rate of change, and c never changes, the rate of change is 0.

Here's another one:

**d/dx(x)=1**

Now, this is telling us that the derivative of x (with respect to x) equals 1. Why is this?

Well, you could represent x as 1x.

1x is the same as 1 * x.

Since it's being multiplied by 1, we could say that the rate of change is 1.

Similarly:

**d/dx(nx)=n**

**This states that the derivative of nx (with respect to x) is n.**

This is because nx represents n times x.

Since we are changing x by multiplying it by n, n is the derivative.

Yes, it is confusing, but eventually you'll get the hang of it

**d/dx(x/y) = {[y * d/dx(x)] - [x * d/dx(y)]} / y^2**

Now this, this is confusing. But it's even more confusing when I put it into words

*"The derivative of x over y is y times the derivative of x, minus x times the derivative of y, all over y squared."*

*Let me give you a demonstration.*

**d/dx(3x****^2/5x) = [(5x * 6x) - (3x^2 * 5)] / 5x^2**

Quickly, here are some other rules:

d/dx(x^n)=nx^n-1

d/dx(1/x)=ln(x)

d/dx(x + y) = d/dx(x) + d/dx(y)

'

And that's all I have for this blog post. Goodbye

]]>Most people learn derivatives with the boring "actual definition", but I learned derivatives (and pretty much all of calculus) with just rules.

Now, let me break this down for you:

You may be asking, why is it 0?

Well, calculus is the study of change.

Derivatives are the rate of change with respect to a variable.

All numbers (1, 3, -8, 3.14159...) that have a value that doesn't change are constants.

Since the derivative is the rate of change, and c never changes, the rate of change is 0.

Here's another one:

Now, this is telling us that the derivative of x (with respect to x) equals 1. Why is this?

Well, you could represent x as 1x.

1x is the same as 1 * x.

Since it's being multiplied by 1, we could say that the rate of change is 1.

Similarly:

This is because nx represents n times x.

Since we are changing x by multiplying it by n, n is the derivative.

Now this, this is confusing. But it's even more confusing when I put it into words

Quickly, here are some other rules:

d/dx(x^n)=nx^n-1

d/dx(1/x)=ln(x)

d/dx(x + y) = d/dx(x) + d/dx(y)

'

And that's all I have for this blog post. Goodbye