If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>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 at ?
• 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."
• 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.
• 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!
• What does g(x) =? I can't read the yellow writing.
• g(x) = |x - 1| + 1
• How did Sal estimate that the slope of 2-x is -1?
• 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.
• At what does he mean by the term 'scalar'? I have done vectors before, but this doesn't make any sense to me.
• 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.
• 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!
• 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.
• How else could we find what is g'(9) except using a graph and with using knowledge that we have learned so far?
• 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.
• At , Sal converts |x-1| to |1-x|. What is the logic? And how did he get 2-x? Please help me out :)
• 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|.
• 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?