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### Course: AP®︎ Calculus AB (2017 edition)>Unit 2

Lesson 2: Estimating derivatives with secant lines and average rate of change

# Estimating derivatives

Estimating derivative at a point using the slope of a secant line connecting points around that point.

## Want to join the conversation?

• f'(4) is not the same as f(4), which I think where you may be getting confused. f'(4) is the derivative/ the slope of the line tangent to the graph at x = 4. 4 is in the middle of 3 and 5, so for the best estimate of f'(4) you would take (y2 - y1) / (x2 - x1) to estimate out f'(4). (112 - 95) / (5 - 3) = 17 / 2 = 8.5. The slope of the tangent line at x = 4 would be 8.5 making f'(4) equal 8.5.

I hope this helps and feel free to comment back with more questions.
• Why don't the question give compatible condition for interpolation ex: Lagrange,Newton interpolation etc?so that we can get the best estimation of it?
• The point is to introduce the concept of numerical estimation of derivatives as secant lines, which is generally the basic concept behind Lagrange interpolation, Newton's method, Euler's method, Taylor approximations, etc.
• Why he didn't draw the tangent line for 3 and 5 depending on the direction of their steepness? why did he draw a line that meets them both even though it's not as per the curve direction?
• We really don't know the curve direction, or the steepness of the curve at any point. All we know for certain is what was given, i.e. a collection of points. The closest points to (4, f(4)) are (3,95) and (5, 112). That is where Sal drew the secant line connecting those points, then calculated the slope of that secant line. This is the best approximation of the slope at (4, f(4))
• so for this type of question, we just generally assume the function is continuous in the related interval?
• Yep! The function is given to be differentiable. And on a particular interval, if a function is differentiable, it is always continuous. There's a proof for this you'll see later on.
• If you DID have the point value for when x = 4, would you still estimate for the graph in the same way (using points around (4, y) instead of (4, y) itself)?
• this is actually easy once you get the hang of it. If it was h'(9), then in the table find the two points closest to 9 in the x values like 8 and 10 or 11. then find the change in y over the change in x and there you go!