 Hi everyone. Let's look at a couple of examples of how to use the two limit definitions of the derivative to answer some questions related to functions, their derivatives, and other follow-up types of questions. So the first one we're going to look at here is the parabola x squared plus 2x plus 1. We are asked to find the derivative y prime at x equals 1. Since we are asked to find that derivative at a specific point, we are going to use the second limit definition of the derivative. We are then asked to find the equations of the tangent line and the normal line. Remember normal in mathematics means perpendicular to the curve at x equals 1. So review of some algebra equation writing skills. So let's start with the calculus part, the derivative. So we are trying to find the derivative and we can just call it f prime of 1 if you wish. So remember this is the one that goes the limit as x approaches 1 of f of x, so the original equation, original function, minus f of c. So in this case we need the function value at 1. We need to find essentially f of 1 on our own and many of our other examples that was given to us in the ordered pair. So we need to substitute 1 into the function and when you do that you do get 4. So this will be minus 4. And our denominator is x minus 1. So as we start to simplify in the numerator we have x squared plus 2x minus 3. In the denominator we are not able to substitute in 1 because it results in a 0 in the denominator. So we are going to have to algebraically manipulate this limit expression by factoring the numerator. That enables the x minus 1s to cancel. So now we can substitute 1 in place of our x and we have 4. Remember what that is, that is the slope of the line tangent to this parabola at x equals 1. So now we can use that to write the equations of the lines we need because you probably and hopefully remember from algebra that in order to write the equation of any line you must have the slope. That is what the derivative provides for us. So let's find the tangent line equation first. So in order to do that we are going to make use of point slope form of a line. Recall that point slope form of a line is y minus y1 equals n times the quantity x minus x1. And all we need to do is substitute in. We have y minus 4 equals 4 which is the slope times x minus 1. We can rearrange that a little bit and we end up with simply y equals 4x because the 4s will cancel. Now let's get the equation of the normal line. Now once again we can use point slope form of a line so we can still start out y minus 4. Now for the slope though remember the slope that will be perpendicular to that which we have here the slope of 4 will be the opposite reciprocal. So the slope of the normal line will be a negative one fourth. And we can simply move the 4 over and add it over if you wish. There is not necessarily a need to distribute everything out. It is quite sufficient to just leave it like that. So let's look at another example. Here we are simply asked to find the derivative of f of x equals 3 times the square root of x minus 3. So since there is no point at which we are trying to find the derivative we will be using the first limit definition of the derivative the one with h in it. So we have f prime of x will be the limit as h approaches 0. So remember the first part of the numerator requires us to do f of x plus h. So in place of that x we need to put x plus h minus f of x all over h. So we obviously cannot substitute 0 in for that denominator so this is one that you hopefully recall we will have to multiply by the conjugate of that numerator expression in order to get this simplified. So the conjugate we will multiply by will be 3 square root of x plus h minus 3 plus 3 square root of x minus 3. Across the numerator we will go ahead and multiply that out. The denominator we shall just leave as it is. So multiplying this out we have the 3 times the 3 in the numerator so that makes 9 times the quantity x plus h minus 3. The outside and inside terms cancel out. In the back we have minus 9 times the quantity x minus 3. And as I mentioned the denominator we shall just copy over and leave as is. So we have some distributing to do in the numerator. Let me go over to the next slide so we have a little bit more room. And things should cancel out nicely if we did this correctly. So when we distribute the 9's in the numerator there we will have 9x plus 9h minus 27 and then minus 9x. So you can see those 9x's will cancel and plus 27 so the 27's will cancel. And that's all over that denominator we had. So these cancelled and these cancelled. The h's then cancelled because all you have there in the numerator is that 9h. So these h's cancel so you have the 9 in the numerator. And then this whole expression with the square roots in the denominator when you substitute in 0 in place of the h here. Remember we had the 9 in the numerator. 3 square root that just becomes x minus 3 because the h was 0 plus 3 square root x minus 3. So that gives you a 6 times square root of x minus 3 in the denominator which of course will simplify in the end to 3 over 2 square root x minus 3. Now that seemed like a whole lot of work to go through to get to the derivative. And soon you shall have some quote unquote shortcuts that will enable you to get to this answer in just one or two steps. And you won't really have to go through all of the algebra as nice as it is that you saw here.