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### Course: AP®︎/College Calculus BC > Unit 2

Lesson 6: Derivative rules: constant, sum, difference, and constant multiple: introduction# Basic derivative rules: table

Let's explore a problem involving two functions, f and g, and their derivatives at specific points. Our goal is to find the derivative of a new function, h(x), which is a combination of these functions: 3f(x)+2g(x). By applying basic derivative rules, we determine the derivative—and thus the slope of the tangent line—of h(x) at x = 9. Created by Sal Khan.

## Want to join the conversation?

- What is d/dx? I know dy/dx is a derivative of a point and the d is a infinitesimally small change in x and y but what does d mean on its own like at0:59?(40 votes)
- You can read d/dx as "the derivative with respect to x." So, for example, you're familiar with seeing:

y = x²

dy/dx = 2x

We can say essentially the same thing without introducing the variable y:

d/dx x² = 2x

In the first case above, we're saying, "given that y = x², the derivative of y with respect to x is 2x. In the second, we're simply saying, "the derivative of x² with respect to x is 2x."(93 votes)

- We are given that h(x)=3f(x)+2g(x) also x=9. Now if i calculate h(9). It comes out to be 3f(9)+2g(9) which is equal to 3+18=21.

Then differentiating h(x) we get h'(x) as 0 because differentiation of a constant is 0.

Please help me out. :)(16 votes)- Hey! Your calculation for h(9) is solid, but there's one problem with your reasoning for h'(x). When you differentiate h, you are not finding the derivative of the concrete value of h(x) (which in your case was h(9)=21). Instead, you are finding the general derivative for the whole function h, and then you plug in your x value of 9 to solve.

So the derivative of h(x) is h'(x)= 3f'(x)+ 2g'(x). Then if we need h'(9), we solve:

h'(9) = 3f'(9) +2g'(9).

Hope this helped!(19 votes)

- What does g(x) =? I can't read the yellow writing.(10 votes)
- g(x) = |x - 1| + 1(7 votes)

- How did Sal estimate that the slope of 2-x is -1?(6 votes)
- y = 2 - x is a straight line. Any straight line with equation y = mx + b has a slope m. That is, the
**slope is the coefficient of x**. So in y = 2 - x, the coefficient of x is -1, so that's the slope of that graph.(17 votes)

- At2:42what does he mean by the term 'scalar'? I have done vectors before, but this doesn't make any sense to me.(4 votes)
- In this context, Sal is using the term scalar to mean a multiplying constant. It is a factor by which the function scales up or down.(14 votes)

- Is there a video covering how to derive an absolute value function on here that I'm just not seeing? Like, I see this and I think I understand, but my professor expects us to show all of our derivatives algebraically, so I'm trying to find a resource to help me learn how to do it (the course is online so I don't have a lecture, but exams are in person). Thanks!(8 votes)
- It might help if you write |x| as sqrt(x^2))and use the chain rule. But at the same time it might be even more simple to use the method kubleeka mentions.(4 votes)

- How else could we find what is g'(9) except using a graph and with using knowledge that we have learned so far?(8 votes)
- Why you don't want to use a graph? Surely, you could write:

if x>1

then f(x)=|x-1|+1=x-1+1=x

therefore f'(x)=1

But that is essentially the same thing as Sal did, without visualisation.

If you are stubborn, you could use the definition of a derivative:

limit as h->0 of (|x+h-1|+1)-(|x-1|+1)/h

But still, to solve that, you would have to divide it to two cases - x>1 and x<1. It would just take more time to solve that.(6 votes)

- At7:13, Sal converts |x-1| to |1-x|. What is the logic? And how did he get 2-x? Please help me out :)(2 votes)
- You can ignore negative signs inside absolute value brackets, since |-x|=|x|. So Sal figured

|x-1|=|-(x-1)|=|-x+1|=|1-x|

Alternatively, you can always interpret |a-b| as 'the distance between a and b'. And then it's obvious that the distance between a and b is the same as the distance between b and a, i.e. |a-b|=|b-a|.(16 votes)

- So, correct me if I am wrong, but what I'm getting right now is that for any given function, we have to find the derivative of the general function FIRST, and then plug in our value if we're given one? Why would it not work if we evaluate the function first for our given 'x'? I've tried it myself and h(9) gives 21, so is there no way to go from there and find the derivative only through the answer?(4 votes)
- If you evaluate the function before taking the derivative, then you'll get a constant number. When you take the derivative afterward (derivative of a constant), it will always be 0, no matter what the function was. And we know it's not true since different functions will have varied slopes (derivatives).

So to find a derivative at a specific x, we first need to find the derivative function then evaluate it.(8 votes)

- What is dy/dx? Is I encountered this in a physics textbook in a section for acceleration and was just wondering...(2 votes)
- dy/dx is the derivative of y with respect to x.(6 votes)

## Video transcript

Voiceover: We've been given some interesting information here about the functions f, g, and h. For f, they tell us for given values of x what f of x is equal to and what f prime of x is equal to. Then they defined g of x for us in terms of this kind of
absolute value expression. Then they define h of x for us, in terms of both f of x and g of x. What we're curious about
is what is the derivative with respect to x, of h of
x at x is equal to nine. I encourage you to pause this video and think about it on your own before I work through it. Let's think about it a little bit. Another way just to get
familiar with the notation of writing this, the derivative of h of x with respect to x at x equals nine. This is equivalent to h,
we need that blue color, it is equivalent to h prime and the prime signifies that
we're taking the derivative. H prime of x, when x equals nine so h prime of nine is what this really is. Actually I'm going to do
this in a different color. This is h prime of nine. Let's think about what that is. Let's take the derivative of both sides of this expression to figure out what the derivative
with respect to x of h is. We get a derivative, I'll
do that same white color. A derivative with respect to x of h of x is going to be equal to the
derivative with respect to x of all of this business. I could actually just,
well I'll just rewrite it. Three times f of x, plus two times g of x. Now this right over here, the derivative of the sum of two terms that's going to be the same thing as the sum of the derivatives
of each of the terms. This is going to be the same thing as the derivative with respect to x of three times, I'll write
that a little bit neater. Three times f of x, plus the
derivative with respect to x of two times g of x. Now the derivative of a number or I guess you could say a
scaling factor times a function. The derivative of a
scalar times the function is the same thing as a scalar times the derivative of the function. What does that mean? Well that just means that this
first term right over here that's going to be equivalent to three times the derivative
with respect to x of f, of our f of x, plus this part over here
is the same thing as two. Okay, make sure I don't
run out of space here, plus two times the
derivative with respect to x. The derivative with
respect to x of g of x. Derivative of h with respect to x is equal to three times the derivative of f with respect to x, plus two times the derivative
of g with respect to x. If we want to write it in this
kind of prime notation here, we could rewrite it as
h prime of x is equal to three times f prime of x, so this part right over here that is the same thing as f prime of x. It's three times f prime of x,
plus two times g prime of x. Once you are more fluent
with this property, the derivative of the sum of two things is the sum of the derivatives. The derivative of a scalar times something is the same thing as a
scalar times the derivative of that something. You really could have
gone straight from here to here, pretty quickly. Now why is this interesting, well now we can evaluate this function when x is equal to nine. H prime of nine is the same thing as three times f prime of nine, plus two times g prime of nine. Now what is f prime of nine? The derivative of our function f when x is equal to nine. Well they tell us, when
x is equal to nine, f of nine is one but more importantly f
prime of nine is three. This part right over here
evaluates that part's three. What's g prime of nine? Let's look at this function
a little bit more closely. There's a couple of ways
we could think about it. Actually let's try to graph it, now I think that could be interesting. Just to visualize what's going on here. Let's say that's our y-axis and do this right over here is our x-axis. Now when does an absolute
value function like this, when is this going to hit a minimum point? Well the absolute value of something is always going to be non-negative. It hits a minimum point when
this thing is equal to zero. Well when is this thing equal to zero? When x equals one, this
thing is equal to zero. We hit a minimum point
when x is equal to one, and when x equals one, this term is zero absolute value of zero, zero. G of one is one. We have this point right over there. Now what happens after that? What happens for x greater than one? Actually let me write this down. G of x is equal to, and in general whenever
you have an absolute value, a relatively simple absolute
value function like this you could think of it,
you could break it up into two function or you could think about this function over different intervals when the absolute value is non-negative and when the absolute value is negative. When the absolute value is non-negative that's when x is greater
than or equal to zero. When the absolute value is non-negative, if you're taking the absolute
value of a non-negative number that is just going to be itself. The absolute value of zero, zero. Absolute value of one is one. The absolute value of
a hundred is a hundred. Then you could ignore the absolute value for x is greater than or equal to, not greater than or equal to zero, for x is greater than or equal to one. X is greater than or equal to one, this thing right over
here is non-negative. It will just evaluate to x minus one. This is going to be x minus one plus one. Which is the same thing as just x, minus one plus one, they just cancel out. Now when this term right
over here is negative and that's going to happen
for x is less than one. Well then the absolute value is going to be the opposite of it. You give me the absolute
value of a negative number that's going to be the opposite. Absolute value of negative
eight is positive eight. It's going to be that the
negative of x minus one is one minus x, plus one. Or we could say two minus x. For x is greater or equal to one, we would look at this expression, now what's the slope of that? Well the slope of that is one. We're going to have a
curve that looks like or a line I guess we could
say that looks like this. For all x is greater than or equal to one. The important thing, remember, we're going to think about
the slope of the tangent line when we think about the derivative of g. Slope is equal to one. For x less than one or our slope now, if we look right over here our slope is negative one. It's going to look like this. It's going to look like that. For the pointing question, if we're thinking about g prime of nine so nine is some place out here, so what is g prime of nine? G prime of nine, let me make it clear, this graph right over here, this is the graph of g of x or we could say y, this is
the graph y equals g of x. Y is equal to g of x. What is g prime of nine? Well that's the slope
when x is equal to nine. The slope is going to be equal to one. G prime of nine is one. What does this evaluate to? This is going to be three times three, so this part right over here is nine plus two times one, plus
two, which is equal to 11. The slope of the tangent line of h when x is equal to nine is 11.