Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Linearization standard linear approximation. If F is differentiable at X equal a, then the approximating function This is a new function L of X equal F of a + F prime of a times the quantity X -. A is the linearization of F at a. In reality this is Y equal Y naughty plus M times the quantity X -, X naughty, which as we know is just point slope formula for a line. We now introduce 2 new variables DX and DY with the property that if the ratio exists, it will be equal to the derivative. So differential means a small change. DF is the differential estimate, while the change of F or triangle F or delta F is the true change. So DF is an estimate, it's a differential estimate, and delta F is the actual true change. DY represents the amount the tangent line rises or falls, whereas delta Y represents the amount that the curve Y equal. F of X rises or falls when X changes by some amount. Let's look at a couple examples. If we look at this 1F of X = sqrt X ^2 + 9 at A equal -4, well, the first thing we need to do is we need to figure out what this point is. So if we're given a is -4, we're just going to stick -4 in for the XF of X equation to find our Y -4 ^2 16 + 916 + 925. Sqrt 25 is five. So the point that we're going to use is -4 five. Now we need to figure out the slope. So we're just going to take the first derivative F prime of X. So the derivative of the inside is 2X. The derivative of the whole thing would be 1 / 2 square roots of X ^2 + 9. The twos cancel, we get X / sqrt X ^2 + 9. When we stick in -4, we get -4 / sqrt -4 ^2 + 9 or negative 4/5. So the linearization is 5 - 4/5 ( X + 4. We're going to distribute it out to put it as a line Y equal MX plus B form. So LX equal -4 fifths X + 9 fifths. This is the linearization of the original function at the point -4 five. If we look at another one here, we're going to have F of X equals sine X, and we want to figure out what happens at X equal π. So the first thing we're going to do is we're going to find our Y value by sticking in π. So F of π or sine of π is just 0. So we know that this is really a point. It's the point PI0. Now we need to find the slope. We find the slope by F prime of X. Well, the derivative of sine X is cosine X, so F prime π is cosine π which is -1. So when we stick it into our formula, we get the linear linearization of 0 - 1 from the slope times the quantity X - π Distributing it out. The linearization is negative X + π. If we do another example, F of X equal the quantity 1 + X to the K at X = 0. So the first thing we're going to do is find F OF0K is a constant here. So F of zero 1 + 0 all to the K power is just one. I'm going to take the derivative of F prime. So we bring the derivative of the inside as one, and then the derivative of the whole thing, bring down the exponent and take it to 1 less power. When we stick 0 in for our X and evaluate, we're going to get out K because anything one to any power is going to be 1, so we're going to get out K. So then linearization is going to be the Y value, which was 01. So one plus the slope of K times the quantity X -, 0 or 1 equal XK. Now we're going to switch examples and we're going to go into differentials. So now we're actually talking about a small change in One Direction of one variable. So when I look at YY is going to change a small amount or the differential dy. When I look at X ^3, X is going to change in terms of the X direction by a small amount. So we're going to take the derivative of X ^3 and get 3X squared, but it's in terms of the X direction. So we're going to put a DX there -3 the derivative of the square root of X is going to be 3 / 2 sqrt X times the differential of X. If you wanted to think about the differential, the change of Y over the change of X, the change of Y over the change of X would really be the derivative. But now we're talking about differentials. We want to look at each direction separately. So the change in the Y direction is equal to three X ^2 -, 3 / 2 square roots of X times the change in the X direction. Those changes, those DYDXS, are called differentials. 2Y to the three. Another example, 2Y to the three halves plus XY minus X all equaling 0. So we're going to do the differential in terms of whichever variable direction we're moving. So for this first 12Y3 halves, we're going to do the differential in terms of Y, bringing down the power to the one less power times DY. This next term we actually have two variables. We have X&Y. So we need to do the differential in terms of X, and then we need to do the differential in terms of Y. So the derivative of X, the differential of X, is going to be 1 DX times the Y, plus the differential of Y is DY times the X minus. The differential of X is just DX. Now we're going to group everything that has a DY together SO3Y to the half plus X. We're going to take everything that doesn't have a DY in it to the other side, so DX minus dxy on the other side. And then we're going to divide by that three Y to the half plus X to get the differential of Y all by itself. Now the fact that we have X&Y on the other side in DX is OK, because eventually we're going to be given a point and we're going to stick the point in the next type of example. We're going to have a couple different pieces and parts to. And the first thing we're going to ask is what's the change or the delta F? So we want the F of X not plus DX minus F of X not. So they give us the equation F of X equal two X ^2 + 4 X -3 and they give us X not as -1 and DX is .1. So the delta F is going to be 2 * -.09 ^2 + 4 * -.09 - 3. This is just my F of X delta plus my differential. Then we're going to subtract my F of X, remembering that X is given as -1 and the delta was .1 so -2 * -1 ^2 + 4 * -1 -, 3. If we go plug plug, chug, chug, we get .02. So the delta change, the actual change of F is going to be .02. Now we're going to do the value of the estimate or we're going to do the differential. So the differential of the function is going to be the first derivative at that location times the differential in terms of X. So going back to the original equation, we're going to take the derivative 4, we're going to take the differentials. So DF equal 4X DX plus 4DX. We're going to stick in the -1 and when we see our DX that was given as .1. So when we do that computation, we're going to get our differential of the function being zero. We want to figure out the approximate error next. The approximate error is the delta of F minus the differential of F the absolute value of that this case it was positive already, so .02 - 0 equal .02. This is the approximate approximation error. Thank you and have a wonderful day.