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# Derivatives of Power Functions - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

To find the equation of a tangent line, first sketch the tangent line. Remember that a tangent line touches the curve of a function at only one point. After sketching the tangent line, find any two points (x_{1}, y_{1}) and (x_{2}, y_{2}) on it. From these two points, you can calculate the slope m (with the usual formula of the change in y divided by the change in x). Then, find the y-intercept of the graph by plugging these points into point-slope form: y-y_{1}=m(x-x_{1}). This will give you the equation of the tangent line.
When you have the equation of the tangent line, you can calculate the derivative of the graph at that point x by plugging in the value of the x you used to draw the tangent line.

Let's take a look at another problem. Suppose I have the function f(x) equals 1 over x² which I have graphed here. First, let's sketch the line tangent to f at x equals -2.

So let me pick another color. Here is x equals -2 right here. I want to sketch the tangent line. One thing that's good about sketching tangent lines ahead of time, is in part b I'm going to find an equation for this line, and having a sketch that will help me make sure that I get the right answer. Looking at this sketch, I can see that I expect a positive slope.

Let's find an equation for the tangent line that I just graphed. So f(x) is 1 over x². That's the same as x to the -2. So the derivative of this f' is -2x to the -3 which is the same as -2 over x³.

Now, to find the equation of tangent line I need two things.; a point, and a slope. For the point at x equals -2, I need a y coordinate f(-2), so that's 1 over -2² or ¼. So my point is -2,¼.

Now what's the slope? The derivative gives me slope. So f'(-2) is -2 over -2³. So that's 2 over -8 which is also ¼, so the slope is ¼.

So in point-slope form, my tangent line is going to be y minus the y coordinate, ¼, equals the slope which is also ¼, times x minus the x coordinate. So x plus 2. So this is an equation of tangent line in Point-slope form.

If you want it slope-intercept form, all you have to do is solve this for y. So you have ¼x. ¼ times 2 is 1/2, and I add ¼ to that and I get 3/4. So y equals 1/4 x plus 3/4.

Let me go back to the graph. That's the equation of line line. 1/4x plus 3/4. The equation of the tangent to f at x equals -2.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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