 ###### 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.

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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.

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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 (x1, y1) and (x2, y2) 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-y1=m(x-x1). 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.