Calculus - Equation Of A Tangent Line Passing Through A Point
Hi everyone. In this math post, I cover the topic of the equation of a tangent line passing through one point of a function. The derivative from calculus is for computing the slope of a function at a certain point.
It is assumed that the reader/student is familiar with calculus derivatives, chain rule, product rule, quotient rule and so on.
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Topics
- Equation Of A Line Review
- Examples
Equation Of A Line Review
From early high school mathematics, the equation of a line is of the form f(x) = mx + b.
where b is the y-intercept of the line, m is the slope of the line, x is the independent variable and f(x) is the dependent variable.
The slope m is the rate of change. Consider a slope of 3 as an example. Every 1 unit increase of x, there is an increase of y of 3 units.
In the calculus framework, the derivative at a point x is the slope of the tangent line passing through the point (x, f(x)). Here is an example of x-squared with a tangent line passing through (1, 1). The tangent line of 2x - 1 passes through the point (1, 1).
Examples
Example One
Determine the equation of a tangent line that pass through x = -1 for the function y = x^3 -1.
Start with the function and take the derivative with the power law.
Once the derivative is determined, substitute x = -1 in the derivative to get the slope of the tangent line.
The equation of the tangent line is of the form y = mx + b. The slope is known which is 3 but we need to determine the (x, y) point corresponding with x = -1. This is x = -1 in the original function f(x).
The equation of the tangent line so far is y = 3x + b
Substitute the point (-1, -2) and solve for the y-intercept b.
The equation of the line that passes through the point x = -1 for the function y = x^3 -1 is y = 3x + 1. Desmos screenshot below.
Example Two
The first example dealt with a simple function. Finding the equation of tangent line passing through a point can get really complex if the function is complicated. Here is one.
Determine the equation of a line passing through the point x = 2 for the function of:
Taking the derivative of f(x) requires the chain rule. You take the derivative of the natural logarithm, multiply it with the derivative of the stuff inside the outer brackets and apply chain rule again for the derivative of 2x -1.
Although the algebra is a bit messy, confusing and technical the derivative ended up being kind of nice & simple. Determine the (x, y) point with x = 2 in the original function f(x) and the (x, y) point with x = 2 for the derivative.
The equation of the tangent line passing through x = 2 for the logarithmic function so far is y = 2x + b.
The slope was determined to be 2 from the derivative at x = 2. Use the point (2, ln(27)) for x and y respectively in the equation of the line & solve for the y-intercept b.
The equation of this tangent line is y = 2x + (ln(27) - 4) or y = 2x - 0.70416. Desmos screenshot below of the log function and the tangent line.
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