Question

Find the derivative of f(x) = 5x + 9 at x = 2.
@sweetsunray

Answer 1

i dont really understand this because it is different from before

Answer 2

what is the derivative f'(x)? This one is easier than the previous one

Answer 3

do i plug in 2?

Answer 4

No, not yet, first take the derivative. Do you know how that's done?

Answer 5

im not sure. im really confused in this lesson and my teacheris on vacation

Answer 6

hello?..

Answer 7

im at the library and my time off of this computer is 4 minutes >.< @sweetsunray

Answer 8

ok, mechanically you need to know 3 things for derivatives of this kind

1) let's say we have a function f(x)=5. This is constant function. De derivative f'(x) of a constant function is always = 0. So in general it is said that if 'a' is a constant and f(x)=a then f'(x)=0

2) If it is of the form f(x)=ax^(b) (that is the variable x with a power multiplied by a constant then the derivative f'(x)=ba x^(b-1)

3) If you have a sum of terms then you take the derivative of each term separately and then sum them up later.

Let's apply these principles to the given function f(x)=5x+9

It is a sum of terms, so we can take the derivatives of each term separately.

Let's start with 5x. You can rewrite it like 5x^1. So if we apply the formula of 2) then a=5 and b=1. The derivative of 5x thus is -> 1.5.x^(1-1) = 5x^(0). If you take power zero of something = 1. So, the derivative of 5x -> 5. You lose the variable.

Now let's do 9. That is a constant. There are no variables there. Actually you could say there is a x^0 -> 9.x^0=9.1=9. Anyhow, it' is a constant. So we can apply the first rule I mentioned. The derivative of 9 = 0. And why is this? Because if you would take the derivative of 9.x^0 you'll get 0.9.x^(-1) = 0.

So, we now have both individual derivatives. Let's sum them up.

f'(x)= 5 + 0 -> f'(x)=5

You can't and don't have to plug in the 2 anymore, because the x is gone. The derivative is a constant function.

Answer 9

wow, that was a perfect explanation. i am copying it into my notes(:

Answer 10

I hope that helped with the basic principles. When you have to take derivatives with x in a denominator or under a squareroot, it's easier to rewrite it as respectively a negative power or a rational power. That way you can see much easier wat b is.

Answer 11

for example 1/x = x^(-1) -> b= -1
other example sqrt (x) = x^(1/2) -> b=1/2