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### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 9: Derivative as a function

# Connecting f and f' graphically

More practice with the relationship between the graph of a function and the graph of its derivative. Created by Sal Khan.

## Want to join the conversation?

• Is it true that if you keep on taking the derivatives of a formula, and graphing the derivative, and then the derivative of that, and then the derivative of that, of a wavy line, the amplitude will continue to decrease and the frequency will continue to increase?
• The effect of repeatedly taking derivatives depends on the nature of the function. Repeatedly taking derivatives of polynomials wipes them out, as the highest order term keeps getting reduced until it becomes a constant and then zero. Repeated derivatives of sine or cosine wrap around in a cycle every four times: sin, cos, -sin, -cos, and back to sin. For hyperbolic sine and cosine the cycle is just two: the derivative of sinh is cosh, and the derivative of cosh is sinh. Most impressive of all, e^x is its own derivative, so that no matter how many times you take the derivative you keep getting the same thing, e^x.
• Sorry that was bad phrasing, my english is not very good. Let me rephrase it. I noticed that when the derivative of y=f'(x) is graphed, it has the same x intercepts as y=f(x). Is this a coincidence or is it a theorem? Can anyone please explain or link me? Thanks! :)
• Here, it's actually just a coincidence. When the second derivative (derivative of the derivative) touches the x-axis, the derivative of the function usually goes from decreasing to increasing or vice versa. In this graph, that just seems to happen at the x-intercepts of f(x). Since at the x-intercepts of f(x), the graph's speed goes from increasing to decreasing or vice versa, the x-intercepts of f(x) and the x-intercepts of the second derivative of f(x) are the same. This is just a coincedence, though, and won't always happen.

I hope this helps!
• Has anyone clearly understood why the green function cannot be f(x) and yellow function, f'(x)?
I think that I repeatedly played the video, but my confusion is not cleared.

Thanks.
• for a function to be f'(x) it should denote the slope of f(x) at that value of x . Consider the part to the extreme left. The slope of the green function is positive but at that value of x the value of the yellow function is negative. So The yellow function cannot be f'(x) . Now, the slope of the yellow function is clearly depicted by the green function so the green function has to be f'(x) and f(x) has to be the yellow function
• at Sal says that the derivative of f(x) should be going downwards or decreasing how did he know that from "eyeballing" it?
I thought the derivative would be positive whenever the function was sloping upwards.u.. but I guess not :(
• Be careful.
You are correct that f '(x) is positive for the whole left side of f(x) where the slope is positive (so say for x values -10 to -4). But f '(x) is only "increasing" from -10 to around -7.5. It's difficult to see on the curve for f(x) but obvious looking at f '(x).

If you were to get the slope of f(x) at the far left it would be increasing 1, 2, 3 and peaking at 4 around x = -7.5. Then decreasing to 3, 2, 1, 0. Note that those decreasing values 3, 2, 1 are still positive. Again its not easy to see just looking at f(x) but the graph of f '(x) makes it clear.

Hope that helps!
• Sal, where can I get the derivative intuition module?
• Do a sreach in the KA searchbar for "derivative intuition module".
• If you equate the derivative to zero, does that mean that the function would change direction?
• Is the slope of the tangent line (derivative) of a point on f(x), the the value of the point on the graph of f '(x) or is it the slope at that point on f '(x)?
• How can you say that the slope of the tangent is 2-half?