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# Derivatives of Power Functions - Problem 3

###### 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 try one more example. Here I have g(x) equals the square root of x. I want to start by sketching the line tangent to g at x equals 9. Now here is a graph of the function g. It's easy enough to draw a tangent line. In part b I'm actually going to find the equation of this line. But it's good to have a graph of the line to show me that the slope should end up positive. It should be a small positive slope.

So in part b I find an equation of this tangent line. First of all, I need both coordinates of the point of tangency. I actually have those right here, but g(9) is root 9 which is 3. So that gives me the coordinates 9,3. So that's the point of tangency.

The only other thing you need to find the equation of a tangent line is the slope. For that we need the derivative. So let's find g'(x). Now g(x) is root x, but that's the same as x to the 1/2. So the derivative is going to be 1/2x to the -1/2. Remember the exponent comes out in front, and you replace it with 1 less.

Now we need the slope, particularly we need the slope at 9. So let me plug in 9, we have 1/2, 9 to the -1/2. 9 to the -1/2 is 1 over the square root of 9, so 1 over 3. 1 over 3 times 1/2 is 1/6. That's the slope of my tangent line. So using point-slope formula, y minus the y value, 3, equals the slope 1/6 times x minus the x value, 9. So that's the equation of the tangent in point slope form.

If you want to write it in slope-intercept form, 1/6x, 1/6 times -9, -3/2. If you add 3 to both sides you get +3/2. Y equals 1/6x plus 3/2. That's the equation of this line. 1/6x plus 3/2. The tangent line to g(x) equals root x at the point 9,3.

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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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