# Slope of a secant line example 1 | Taking derivatives | Differential Calculus | Khan Academy | Summary and Q&A

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August 2, 2013
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Slope of a secant line example 1 | Taking derivatives | Differential Calculus | Khan Academy

## TL;DR

This video explains how to find the slope of a secant line for a curve with the equation y = ln(x), using the points P(e,1) and Q(x, ln(x)).

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### Q: What is the equation of the curve discussed in the video?

The curve has the equation y = ln(x).

### Q: What are the coordinates of point P?

Point P has the coordinates (e, 1).

### Q: What is the change in x between points P and Q?

The change in x is x - e.

### Q: How is the slope of the secant line calculated?

The slope of the secant line is calculated by taking the change in y (ln(x) - 1) divided by the change in x (x - e).

## Summary & Key Takeaways

• The video discusses finding the slope of the secant line of a curve with the equation y = ln(x) using the points P(e,1) and Q(x, ln(x)).

• The natural logarithm of x approaches negative infinity as x gets smaller, and the curve passes through the points P and Q.

• To find the slope of the secant line, the change in y (ln(x) - 1) is divided by the change in x (x - e).