# Derivative Properties and Formulas – Secrets Exposed!

## Derivative Properties and Formulas

Hey guys. In this video we will take a look at a couple key properties and formulas that will help you find **derivatives** in a much simpler way, rather than using the definition every time.

The first property we will look at is differentiating a __sum or a difference.__

**Here is the property:**

(f(x) + g(x))’ = f’(x) + g’(x) or we could write this in fraction notation d/dx (f(x) + g(x)) = df/dx + dg/dx.

d/dx just signifies that you are taking the derivative with respect to x. So, over on the right here (point) we can see that we have taken the derivative of f with respect to x and the derivative of g with respect to x.

What this property is telling us is that when differentiating a function that has two terms being added or subtracted, you can take the derivative of the terms individually, and simply put them back together with the correct sign.

The next property we will look at deals with __factoring out a constant.__

(cf(x))’=cf’(x)

Or again, we can right this in fraction notation d/dx (cf(x))= c(df/dx).

What this property is telling us is that we can take this constant that is being multiplied by everything in the function, and factor it out. So, then we would just take the derivative of the function, then move the constant back in front.

Now, before we go on to look at some examples of these properties being applied, I must first equip you with some derivative formulas so that you can actually find the derivative.

Here is the first one:

If f(x)=c then f’(x) = 0

OR d/dx (c) = 0

This formula shows us that the derivative of any constant is just zero.

If f(x) = x^n then f’(x) = nx^n-1

Or d/dx (x^n) = nx^n-1 where n is any number.

What this formula tells us to do is to take our exponent and bring it down and multiply it, then subtract one from the exponent.

The n in our formula here has to be a number, not a variable. The base number however, must be a variable to be able to use the power rule or else it would just be a constant.. Which we know is equal to zero from our first formula.

Alright, now let’s take a look at a few examples.

**Differentiate the following function.**

f(x)= 6x^24 – 8x^7 + 7x + 8

So, according to our first property we are able to take the derivative of each term individually, because each term is being added or subtracted. By looking at our function we can see that we will need to use the power rule on the three terms that have **exponents** and a variable in the base, and we know that 8 will be 0, because it’s a constant.

When we apply our power rule, we bring down our exponents to multiply, then subtract one from our exponent.

f’(x)= 6(24)x^23 – 8(7)x^6 + 7(1)x^0 + C

Any time you take the derivative of a constant, you always replace it with + C. Now, we can go ahead and simplify this even further.

f’(x) = 144x^23- 56x^6 + 7 + C

The reason I wrote out 7 times 1 times x0 is to show you why 7x just becomes 7. We know that 7x just has an exponent of 1. So, when we bring that exponent down to multiply which 7 times 1 is just 7, then when we subtract 1 from 1 we are left with zero in the exponent place.. And anything to the power of zero is just 1. So, then that is 7 times 1 times 1 which is just 7. This is true for any variable with an exponent of one. You can just drop the variable and the derivative is just the remaining coefficient.

Let’s look at another example.

**Differentiate the following function.**

g(t)= 1/t^80 + 1/18t^12 + t – 24

This may seem a little scary. If you are anything like me, you are thinking why on earth are there fractions, and you didn’t tell us what to do when we have a fraction. That’s okay, and you are right I haven’t told you what to do yet.. That is another video.

However, we can rewrite this to get out of having to deal with fractions. If you recall the exponent rule that states 1/an= a-n. So, with this in mind we can go ahead and rewrite our function.

g(t) = t^-80 + 18t^-12 + t – 24

Now, we can use our power rule, and formula for a constant.

g’(t)= -80t^-81 -216t^-13 + 1 + C

The reason our exponent here went from -80 to -81 is because we are subtracting 1 from -80. Which, gives us -81, and same thing for our -13 here.

I hope this video on derivative properties and formulas has been helpful to you. For further help be sure to check out more of our videos by subscribing to our channel below.

See you next time!