[0:00]In this lesson, we're going to focus on how to find the equation of a tangent line given a point using derivatives. So, let's say if we have a function, f of x and it's equal to 2x squared minus 5x plus 3. And we wish to find the equation of the tangent line when x is equal to 2. So, what equation should we use? We need to use the point-slope form of a linear equation. At least that's the preferred form I like to use. You can also use the slope-intercept form if you prefer it that way. Now, we need to find two things. Right now we already have x one. X one is two. We need to calculate y one, that is y when x is two and we need to determine the slope. So, to find the y coordinate, we simply have to take x and plug it into this equation. Make sure you understand this. f of a is equal to y, where a is a specific value of x. In this example, a is equal to two. Now, x could be anything. x could be five, seven, twelve. A is that specific value of x. So, f of a is y, and f prime of a is equal to m. So, in order to find m, you have to find the first derivative and then replace x with two. In order to calculate y, you need to replace x with two in the original function, f of x. So, let's determine y first. So, y is equal to f of 2. This is y one specifically. And that's going to be 2 times 2 squared minus 5 times 2 plus 3. 2 squared is 4, 5 times 2 is 10, and 2 times 4 is 8. 8 minus 10 is negative 2, negative 2 plus 3 is 1. So, y one is 1. Now, the only thing we need to do is calculate the slope when x is equal to 2.
[2:21]So, first, we got to find the first derivative. The derivative of x squared is 2x. The derivative of x is 1. And the derivative of a constant like 3 is 0. So, f prime of x is equal to 4x minus 5. The slope is going to equal f prime of 2. So, that's going to be 4 times 2 minus 5. 4 times 2 is 8, 8 minus 5 is 3. So, the slope of the tangent line is equal to 3 when x is 2. Now, make sure you understand the difference between the derivative and the slope. The derivative is a function in terms of x in this particular example. The slope is a specific value of the derivative function at some x value. So, when x is 2, the slope is 3. However, the slope can change with this function. For example, when x is 1, the slope will be negative 1. So, the derivative is a function that gives you the slope at any x value.
[3:35]Now, let's plug in what we have so that we can write the equation of the tangent line. So, it's y minus 1, which is equal to 3 times x minus 2. So, this is the answer in point-slope form. But let's get it in slope-intercept form. So, let's solve for y. Let's distribute the 3 first. So, it's going to be 3x minus 6. And then let's add 1 to both sides. Negative 6 plus 1 is negative 5.
[4:08]So, here we have the equation of the tangent line in slope-intercept form. Let's work on another example. So, let's say that f of x is equal to 8x minus x squared. Write the equation of the tangent line at x equals 7. So, go ahead and work on this problem. Feel free to pause it and see if you got the concept. So, the first thing I'm going to do is calculate the y value. So, let's call this x one. Y one is going to equal to f of 7, which is going to be 8 times 7 minus 7 squared. 8 times 7 is 56. 7 times 7 is 49. 56 minus 49 is 7. So, y one is 7. So, now that we have x and we have y, all we need to find is m.
[5:14]So, let's go ahead and determine the first derivative of the function. So, f prime of x is going to equal the derivative of x is 1. The derivative of x squared is 2x. So, it's 8 minus 2x. Now, let's determine the slope, which is going to be the first derivative evaluated at an x value of 7. So, it's f prime of 7. So, it's going to be 8 minus 2 times 7. 2 times 7 is 14 and 8 minus 14 is negative 6. So, the slope is equal to negative 6.
[6:02]Now, let's use the point-slope formula. Y one is 7, m is negative 6, x one is 7. So, that's the equation of the tangent line in point-slope form, which you can leave it like that if you want to. But let's get it in slope-intercept form. So, I'm going to distribute the negative 6. So, it's going to be negative 6x, and then negative 6 times negative 7, that's positive 42. Now, we need to add 7 to both sides. Negative 6x plus 49. This is the equation of the tangent line in slope-intercept form. That's the final answer. Now, let's work on one more example. But this time the example is going to be associated with a trig function. So, let's say that f of x is 4 sin x minus 3. And we wish to write the equation of the tangent line when x is equal to pi over 6. Go ahead and work on that example. Now, first, let's calculate y when x is pi over 6. So, we need to evaluate the function at pi over 6. So, this is going to be 4 sine pi over 6 minus 3. Now, what is sin pi over 6 equal to? Pi is equal to 180 degrees. So, 180 divided by 6 is 30. So, pi over 6 in radians is equivalent to 30 degrees. Sine 30 or sin pi over 6 is 1/2. And you can use the unit circle or your calculator to figure that out. 4 times 1/2 is 2. 2 minus 3 is negative 1. So, therefore, y is negative 1 when x is pi over 6.
[8:06]Now, let's calculate the slope when x is pi over 6. So, we need to find the first derivative. So, f prime of x is going to equal, so, what is the derivative of 4 sine x minus 3? The derivative of sine is cosine, and the derivative of negative 3 is 0. So, it's just 4 cosine x. So, now let's evaluate it when x is pi over 6. So, this is going to be 4 times cosine pi over 6. Now, what is cosine pi over 6 equal to? What's cosine of 30 degrees? Cosine pi over 6 is equal to the square root of 3 over 2. Now, hopefully, you remember your 30 60 90 triangle. This can help you with these things. Across the 30 is 1, across the 60 is the square root of 3, across the 90 is 2. So, let's say if you want to evaluate sine 30, using so, you know that sine is opposite over hypotenuse. So, opposite to 30 is 1, and the hypotenuse is across the box, which is the longest side, that's 2. So, sine 30 or sine pi over 6 is 1/2. Cosine 30, based on so, cosine is adjacent over hypotenuse.
[9:29]Adjacent to 30 is the square root of 3. And the hypotenuse is still 2. So, that's a quick review that can help you to remember some basic trig functions, whenever you need to evaluate it. Now, let's finish this problem. 4 divided by 2 is 2. So, the slope is equal to 2 square root 3.
[10:04]Now, let's go to the point-slope formula of a linear equation. Y one is negative 1, m is 2 square root 3, and x one is pi over 6. Now, y minus negative 1 is the same as y plus 1. So, you can leave it in this form if you want to. In fact, I think it's best to leave it in this form. You don't want to multiply 2 root 3 by pi over 6. It's not going to look nice. So, I'm going to leave the answer in point-slope form. So, that's the equation of the tangent line for this problem.
