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## Differential Calculus (2017 edition)

### Course: Differential Calculus (2017 edition)>Unit 5

Lesson 1: Introduction to differential calculus

# Derivative as slope of curve

AP.CALC:
CHA‑2 (EU)
,
CHA‑2.B (LO)
,
CHA‑2.B.3 (EK)
,
CHA‑2.B.4 (EK)
,
CHA‑2.C (LO)
,
CHA‑2.C.1 (EK)
Sal solves a couple of problems where he interprets the derivative of a function at a point as the slope of the curve, or of the line tangent to the curve, at that point.

## Want to join the conversation?

• I'm quite confused, the derivatives gotten above are depending on how long Sal drew his slope
• we don't care about how long is that line, we care about its slop (how steep or shallow is it) because for a line you can take any two points (now matter how far or how close) on it and you will have the same slop since it's a line.
and you can check it at :
Sal took the points: (5,5) and (6,7)
so the slop is (7-5)/(6-5)=2/1=2
take another points for the same line for example:
(4,3) and (6,7)
so the slop (7-3)/(6-4)=4/2=2
which is THE SAME SLOP since it's THE SAME LINE.
• I do not get the concept and the definition of the derivative. What exactly is a derivative?
• Suppose you have a function f(x). Its derivative f'(x) describes the instantaneous rate of change of f(x) for any x in the domain. Suppose I told you that f(3)=7. Now you know where the function is at x=3, but you know nothing of its motion. Is it increasing? Decreasing? How quickly. If I tell you that f'(x)=10, that would indicate that at x=3, f(x) is increasing quickly. The intuition will become easier as you keep studying calculus. It might be worth checking out the videos on the relationship between limits and derivatives.
• Around the answer to the second example is that g'(4) > g'(6). I guess it makes sense as g'(4) is less negative than g'(6)... but I find this confusing because the instantaneous rate of change at g'(6) is greater than at g'(4), and the example would make it seem like those instantaneous rates of change are what's being compared. Am I off base here?
• it's kind of similar to comparing velocity and speed(absolute value of velocity).
g'(6) has more "speed" than g'(4), but g'(4) and g'(6) are both negative, so g'(4) has a higher velocity than g'(6) because it's less negative.
i hope i haven't confused you.
• Why can't we differentiate acceleration again with respect to time?
• The derivative of acceleration with respect to time is a quantity called jerk, probably because the greater the absolute value of the rate of change in acceleration, the more "jerky" the motion is.
• On the first problem, the derivative of the constant turned out to be 2. But i have also heard that the derivative of a constant is always zero [so the d/dx (5) = 0]. Im really confused someone please help

Thanks
• For the first problem, we're estimating the slope of f(x). If it were constant, the given graph would be a horizontal line. What might have thrown you off is that we're estimating the derivative at a single point. When people say that the derivative of a constant is zero, the "constant" is a function such that f(x)=c. Taking the derivative at a single point, which is done in the first problem, is a different matter entirely. In the video, we're looking at the slope/derivative of f(x) at x=5. If f(x) were horizontal, than the derivative would be zero. Since it isn't, that indicates that we have a nonzero derivative.
• Can someone explain where these tangent lines are coming from? What is their purpose?
• At , Sal says that g'(4) is > than g'(6) which does not make sense considering that g'(6) is much steeper than g'(4) and if you take the absolute values of both -1 and -3, you get |-1| < |-3|(It does not matter whether the slope is positive or negative. When it comes to the slope of a line, all that matters is its absolute value). So shouldn't the correct answer be g'(4) < g'(6)?.
• In math, a slope of a function is always considered from left to right, which gives us positive or negative slope. So it matters if the slope is negative or positive. It's true that their absolute values would have different effect but you cannot just do that. It's true the steeper the greater when you don't take negative and positive into consideration, but in math, we have to consider them as negative or positive.

g'(6) is much steeper comparing to g'(4), but there are both negative. Because of that g'(6) < g'(4)
• is there a way to figure out the derivative without just eyeballing it? like when the two derivatives are real close to compare
• Well there are derivate rules that you can use to get to the derivative. Getting the derivative by eyeballing the function only really works as long as you work with very simple functions.