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# The Definition of the Derivative - 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 estimate the derivative of a function, you can use the definition of the derivative, which can be written as the limit as h goes to 0 of (f(a+h)-f(a))/h. First, evaluate the function at a. Then, you can calculate (f(a+h)-f(a))/h for various values of h, as h gets smaller and smaller. (It may be easier to evaluate the derivative by simplifying the expression before plugging in values of h). As h gets smaller, it approaches 0 without ever actually being equal to 0. This expression will get closer and closer to some number, which we say is an estimate of the derivative of a function.

Let's try a harder problem that involves the definition of the derivative. Part A says find the derivative of f(x) equals 1 over x at x equals -2. So the first thing I want to do is compute the difference quotient and simplify it. For this function, the difference quotient is going to be f of; a is -2 plus h minus f(-2) over h. Now f is the function of 1 over x. So this will be 1 over -2 plus h, minus 1 over -2. All that over h.

So let me first simplify this expression. I'll take it up here. Now the way to simplify this, is to multiply the top and bottom of this fraction, by something that gets rid of the denominators of the little fractions. So, I'm going to multiply the top and bottom by (-2), and (-2 + h). What that does is, it gets rid of these denominators. So for example 1 over -2 plus h, times this quantity. The (-2 plus h) cancel, and I'm left with -2. Minus, from this minus, this fraction 1 over -2 times this quantity. The -2s cancel, and I get (-2 plus h). I'll put that in parenthesis, because I have to remember to distribute that minus sign.

Then in the denominator I get h times -2, times -2 plus h. When I'm simplifying difference quotients, I like to leave things factored, because it may turn out that it's better that way.

Now in the numerator when I distribute this minus sign, I get -2 plus 2, and they cancel. I get minus h over h times -2 times (-2 plus h). The h's cancel. So I'm left with -1 over -2 times -2 plus h. So this is my difference quotient simplified.

Now that it's simplified, I can find my derivative, which just means taking the limit of this expression. -1 over -2 times (-2 plus h). Now what happens to this expression as h goes to 0? I have -2 times -2 plus 0 in the denominator, 4. So it's -1 over 4. -¼, that's my derivative of the -2. That means that's the slope of the tangent line.

Now we're in luck. We're asked to find the equation of the tangent line, to this function at x equals, this should be -2. So whenever you want to find the equation of a line, you need two things. You need a point, and a slope. The point is -2,-½. The slope is -¼. So we use the point slope formula; y minus the y coordinate, which is -½, so y plus ½ equals -¼x minus the x coordinate, which is x plus 2. That right there is a perfectly good equation for your tangent line.

But your teacher may want in slope intercept form. If that's the case, you want -¼x, 2 times -¼ is -½, and you subtract 1/2 from both sides, and you get -1. So y equals -¼x minus 1. When we draw this tangent line, it looks something like this. Perfect, it even goes through -1 like it's supposed to. So you've got a slope of -¼, and a y intercept of -1. This is the tangent line for the graph of 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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