Here’s an amazing fact: The derivative in ~ a point is the **slope of the tangent line** in ~ that allude on the graph.

You are watching: Finding the slope of a tangent line at a given point

Jenn, Founder historicsweetsballroom.com®, 15+ Years suffer (Licensed & Certified Teacher)

Why is this fascinating?

Because if us are ever asked to solve troubles involving the slope of a tangent line, every we require are the same an abilities we learned earlier in algebra for creating equations that lines.

## Equations the Lines

So, what perform we remember around equations because that lines?

Well, lock require simply two elements:

PointSlope### Example

For instance, if we wanted to compose the equation that a line provided the point (6,1) and slope m = 3. All we will do is substitute the provided information into the point-slope formula and also simplify, as suggested below.

eginequationeginarrayly-y_1=mleft(x-x_1 ight) ext if (6,1) ext and m=3 \y-1=3(x-6) \y=3 x-17endarrayendequation

## Equation that Tangent Line

This method that to uncover the equation that a tangent heat to a curve, f(x), we simply need two elements: suggest and slope. The only distinction is that to discover our steep (i.e., price of change), us will use **derivatives**!

Is her mind blown yet?

### Example

Alright, expect we space asked to create the equation that the line tangent to the curve (y=x^2 ext in ~ x=3).

First, we will discover our point by substituting x = 3 right into our duty to recognize the matching y-value.eginequationeginarraylf(x)=x^2 quad x=3 \f(3)=(3)^2=9 \(3,9)endarrayendequation

Next, we take the derivative of our curve to find the rate of change.eginequationf^prime(x)=2 xendequation

We will certainly then swap our provided x-value into our derivative to discover the slope at x = 3.eginequationf^prime(3)=2(3)=6endequation

Lastly, we will certainly substitute our allude (3,9) and slope m = 6 right into the formula for point-slope kind and write the equation that the tangent line.eginequationeginarrayly-y_1=mleft(x-x_1 ight) ext if (3,9) ext and m=6 \y-9=6(x-3) \y=6 x-9endarrayendequation

See, detect the equation that the tangent line is easy!

We were able to usage our algebra skills to discover the equation of the heat tangent to a curve.

Find The Equation that The Tangent Line

Therefore, stop formally lay the end the measures for composing the tangent line equation come a curve, as this details skill is pivotal for future lessons managing linearization and differentials.

Substitute the provided x-value right into the function to find the y-value or point.Calculate the first derivative the f(x).Plug the ordered pair into the derivative to discover the slope at the point.Substitute both the allude and the slope from measures 1 and also 3 into point-slope type to find the equation because that the tangent line.## Normal heat Equation

Likewise, we can even extend this ide to composing equations of typical lines, which are likewise called perpendicular lines. The only difference will be that us will merely use the an unfavorable reciprocal steep of the heat tangent.

### Example

For this problem, consider the curve (f(x)=2^3 x). Discover the tangent heat equation and normal heat to f(x) in ~ x = 1.

First, we will find our point by substituting x = 1 right into our function to determine the corresponding y-value.eginequationeginaligned&f(x)=2^3 x quad x=1\&f(1)=2^3(1)=8\&(1,8)endalignedendequation

Next, we take the derivative that f(x) to discover the price of change.eginequationf^prime(x)=2^3 x cdot ln (2) cdot 3endequation

Next, we will swap our offered x-value right into our derivative to uncover the slope at x = 1.See more: 72 Is What Percent Of 60 Is 72 As A Percentage? 72 Out Of 60 Is What Percent

eginequationf^prime(1)=2^3(1) cdot ln (2) cdot 3=24 ln 2 approx 16.64endequation

This means that the steep of the tangent line is 16.64, and also the slope of the common line is -1/16.64 or -0.06, i beg your pardon is the an unfavorable reciprocal slope!

Lastly, we will certainly write the equation of the tangent line and normal lines utilizing the allude (1,8) and slope tangent steep of m = 16.64 and also normal slope of -0.06, respectively.

Find The Equations the The Tangent and also Normal present Of The Curve because that A given Point

Simple!

Together we will certainly walk through three examples and learn just how to use the point-slope type to create the equation the tangent lines and normal lines.