# Equation of a tangent line | Taking derivatives | Differential Calculus | Khan Academy | Summary and Q&A

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July 3, 2008
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Equation of a tangent line | Taking derivatives | Differential Calculus | Khan Academy

## TL;DR

Find the equation of the tangent line to the function f(x) = xe^x at x=1.

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### Q: What is the derivative of the function f(x) = xe^x at x=1?

The derivative, or slope, at x=1 is 2e.

### Q: How is the equation of the tangent line obtained using the derivative and a known point on the curve?

By substituting the slope (2e) and the coordinates of the point (1, e) into the equation y = mx + b, the y-intercept can be solved for, resulting in the equation y = 2ex - e.

### Q: Can the equation be simplified without using the constant e?

Yes, by approximating its value as 2.7, the equation becomes y = 5.4x - 2.7.

### Q: How is the validity of the obtained equation confirmed?

By graphing the equation of the tangent line and the original function, it can be visually verified that the tangent line intersects the curve at (1, e).

## Summary & Key Takeaways

• The video demonstrates how to find the equation of a tangent line to a curve at a given point.

• The function f(x) = xe^x, graphed using a calculator, represents the curve in question.

• The point at which the tangent line is sought is (1, e), and the goal is to determine the equation of the tangent line.