in the second part of section 3.2 we continue with a discussion of the derivative as a function so first off here i'd like to find the derivative of f of x equals the square root of 2 minus 2x so as we discussed in the previous video the derivative f prime of x is defined as the limit as h goes to zero of our difference quotient f of x plus h minus f of x all divided by h so filling in the details what we have is the limit as h goes to zero so f of x plus h is going to be x plus h plugged into our function f so i'll have 2 minus 2 times x plus h and then minus f of x which is just minus the function 2 minus 2x and then that's all divided by h so what you might notice here is that direct substitution of zero gives us zero over zero so the presence of the square root might suggest that we multiply by the conjugate so that's going to be 2 minus 2 times x plus h plus the root of 2 minus 2x and then the same thing on the bottom so we'll have 2 minus 2 times x plus h under the root plus the root of 2 minus 2 x so continuing forward we can now begin to simplify our expression a little bit so the conjugates multiplying on top again the front two terms will multiply so i'll drop the root i'll have two minus two times x plus h and then the last two terms will multiply so these two roots multiplying will drop the root and i'll have minus 2 minus 2x and that's all going to be divided by h so continuing this is going to be the limit as h goes to 0 of 2 minus 2x minus 2h now let me back up a little bit i forgot the conjugate on the bottom so this is h times the root 2 minus 2 times x plus h plus the root 2 minus 2x so the conjugate multiplies across on the bottom so again the top will continue as we were performing so this 2 will distribute and i'll get minus 2x minus 2h the negative will distribute i'll get minus 2 plus 2x that's all going to be divided by h times this conjugate of 2 minus 2 times x plus h plus the root of 2 minus 2x now on the top 2 minus 2 is 0 negative 2x plus negative 2x is also 0. so on the top i'm just left with negative 2h so this is going to continue so we'll have this being equal to the limit as h goes to zero we just have a negative two h on the top on the bottom we have h times this conjugate two minus two times x plus h plus the root of two minus two x now notice the two so excuse me the h's are going to drop out so we have now the limit as h goes to zero on the top i have a negative two on the bottom we just have the conjugate two minus two times x plus h under the root plus the root two minus two x so as h goes to zero what we can do is we can substitute in zero for h here that's gonna make this vanish all right so this h here is gonna go to zero so what i'm left with is just a negative two on the top on the bottom i'm going to have root 2 minus 2x plus root 2 minus 2x okay so that's going to be negative 2. now i have a pair of those root 2 of 2 minus 2 x's so i have two root two minus two x these twos are going to drop out and so we have our derivative our derivative is going to be negative one over the root of two minus two 2x so let's go through this process one more time so we want to find the derivative of f of x equals x over x plus 1. so again the definition says that the derivative f prime of x is equal to the limit as h goes to zero and we're going to have f of x plus h the difference quotient minus f of x divided by h so what we have is the limit as h goes to zero so f of x plus h is going to be x plus h over x plus h plus one minus f of x which is just x over x plus one that's all going to get divided by h so we have the limit as h goes to zero so we need to combine these two fractions so what we're going to have is we're going to have x plus 1 times x plus h minus x times x plus h plus one divided by we'll have x plus h plus one times x plus one on the bottom and we're dividing by h which is the same as multiplying by 1 over h so we'll continue working through this now perhaps what you should notice here you know this is certainly calculus but 95 percent of this is algebra so the only thing that's really new here is this idea of the limit manipulating an expression like this is something that we are able to do simply from knowing our algebra so this is really just one big algebra problem to me with this concept of the limit tacked on to it so we need to simplify some things before we can move forward here i have a product of two binomials so i'd like to foil that that's going to be x squared plus xh plus x plus h and then now the second part will distribute this negative x so we'll have negative x squared minus x h minus x now the bottom we're going to leave as is there's no need to distribute and multiply these two terms together and this is all times 1 over h or dividing by h now notice some good things happen these x squares drop these x h terms drop and so do these x terms so in the top all we're going to be left with is an h so we'll continue this this is now going to be the limit as h goes to 0 and so what we're going to have is just an h left over on the top on the bottom we have x plus h plus 1 times x plus 1 multiplied by 1 over h and so clearly these h's are going to drop out so our expression is now just the limit as h goes to 0. what we have left over on the top is just a one divided by x plus h plus one times x plus one substitution of zero is going to give us one over x plus one times x plus one which we can write as 1 over x plus 1 squared so the derivative of this function f of x equals x over x plus 1 is going to be f prime of x equals 1 over x plus 1 quantity squared so we have a connection here between the idea of differentiability and continuity and the statement is if a function is differentiable at some value x equals c then it will also be continuous at that same value so what we might say is that differentiability implies continuity however continuity does not imply differentiability so if you consider the absolute value function so the absolute value function has this general shape it's piecewise it joins the two lines of y equals x and y equals negative x together at the origin so where the two lines meet here at zero so on the right hand side our derivative is going to be one so again this is the line y equals x so the slope of that line is one so we find that the derivative as we approach zero from the right is one and as we approach zero from the left we find the derivative equal to negative one so the derivative being a two-sided limit one is of course not equal to negative one so for the absolute value function f prime of zero does not exist even though we are continuous right here at the value of zero so differentiability guarantees continuity but continuity does not guarantee differentiability so in general when will a function fail to be differentiable we have a handful of cases here so first any discontinuity will cause the derivative to not exist second the existence of a corner as we've seen with the absolute value function or a cusp will also cause the derivative to not exist so a cusp has sort of this general shape to it so again we have a very rapid change in direction so at these types of places on the graph of a function we would not have a derivative here at this point at the top third the existence of a vertical tangent will cause the derivative to not exist again if we have a vertical tangent what we have is effectively an infinite slope and of course limits that are equal quote-unquote to infinity do not exist and then lastly the derivative will not exist at an end point since a two-sided limit is not possible at an endpoint and we really don't talk about the ideas of left or right differentiability although we could certainly define those things if we wanted to so these four cases any discontinuity a corner or a cusp a vertical tangent and an endpoint we will not have a derivative at these types of locations so in this graph here i want to circle the locations on the closed interval negative 2 to positive 2 where this function graphed is not going to be differentiable in other words where the derivative is not going to exist so first of all we see this absolute value type corner so we will not have a derivative at negative one so no derivative at negative one so we're not going to have a derivative at zero so let's give a justification we've got a corner at zero we've got a jump discontinuity right here at one we have a removable discontinuity and at negative 2 we have an end point no derivative and at positive 2 we also have an endpoint so no derivative so for this graph these five locations we will not have a derivative existing at these points