6x 2 8x + 5 ) = 12x 8. f (x) ) = d (12x 8) = 12

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AMS/ECON 11A Class Notes 11/6/17 UCSC *) Higher order derivatives Example. If f = x 3 x + 5x + 1, then f = 6x 8x + 5 Observation: f is also a differentiable function... d f ) = d 6x 8x + 5 ) = 1x 8 dx dx The derivative of f is called the second derivative of f, and is denoted by f. Observation: f is also a differentiable function... d f ) = d 1x 8) = 1 dx dx The derivative of f is called the third derivative of f, and is denoted by f. And so on... The derivatives of the derivatives of the derivatives, etc.) of y = f are called the higher order derivatives of f. Notation: We use the following notation interchangeably for derivatives and higher order derivatives of a function y = f First) derivative: y = f = dy dx = d dx f) Second derivative: Third derivative: Fourth derivative: y = f = d y dx = d dx f) y = f = d3 y dx 3 = d3 dx 3 f) y ) = f ) = d y dx = d dx f).. n th derivative: y n) = f n) = dn y dx = dn n dx f) n Observation: For derivatives of order or higher, people generally do not write expressions like For obvious reasons. f 17 Example. Find the third derivative of the function w = u. First, write w = u 1/, then dw du = 1 u 1/ = d w du = 1 u 3/ = d3 w du 3 1 = 3 8 u 5/

Example. Find the second derivative of f = xe x. f = product rule 1 e x + xe x }{{} chain rule f = e x + xe x }{{} f = f = e x + e x + xe x ) = e x + xe x Comment: The derivative f gives the rate of change of y = f with respect to x, and this gives useful information about the function. Geometrically for example, as we have seen, f x 0 ) is the slope of the line tangent to the curve y = f at the point x 0, fx 0 )). Likewise, f x 0 ) is the slope coefficient of the linear function T 1 = fx 0 ) + f x 0 )x x 0 ) that we use in the linear approximation of f. This leads to the... Question: How can higher order derivatives be used? *) We will see that the second derivative f has an observable geometric interpretation, describing an important aspect of the shape of the graph of y = f. *) Another way we can use higher order derivatives is to improve on linear approximation. *) Taylor polynomials. To improve on linear approximation, we use polynomials of degrees, 3, etc. to replace the linear polynomials T 1. Example. The graph of the function f = x + x + 5 black), and the graph of the linear approximation to this graph at the point, 3), T 1 = 3 0.x ) blue), are both displayed in the figure below. 1 5.6 y=t.8 y=t 1 3. y=f x). 1.6 0.8.5 0 0.5 1 1.5.5 3 3.5.5 5 5.5 6 6.5 7 Figure 1: Linear approximation to f = x + x + 5, centered at, 3). 1 This example is taken from the supplementary note on Taylor polynomials SN 7) on the course website.

Observations: 1. The linear approximation is good when x, say 3.5 < x <.5.. Linear approximation becomes much less accurate as x moves away from. Why? One explanation is that the graph of y = T 1 is a straight line with constant slope, but the slope of y = f is changing, and as x moves away from, the graph y = f bends away from the graph y = T 1. In other words, the linear approximation f T 1 is reasonably good when x because T 1 ) = f) and T 1) = f ), but less accurate when x moves away from because the slope of T 1 is not changing like the slope of f Idea: To improve on linear approximation, find a function T with the following properties. i) T ) = f) ii) T ) = f ) iii) T ) = f ) Furthermore, T should be as simple a function as possible. *) A linear function won t work here, because if T is linear, then T = 0 but f ) 0. *) The next simplest type of function is quadratic, so we try something like T = A + Bx ) + Cx ). Using x ) instead of x makes the algebra a little easier.) Condition i): T ) = f) = A + B ) + C ) = f) = A = f). Condition ii): T = B + Cx ), so Condition iii): T = C, so T ) = f) = B + C ) = f ) = B = f ) T ) = f ) = C = f ) = C = f ) Conclusion: T = f) + f )x ) + f ) x ) Calculations: f = x + x + 5 ) 1/ = f) = 3 f = 1 x + x + 5 ) 1/ x+) = x) x + x + 5 ) 1/ = f ) = 0. 3

f = 1) x + x + 5 ) 1/ + x) 1 x ) + x + 5 ) 3/ x) [ x = + x + 5 ) 1/ + x) x + x + 5 ) ] 3/ Final conclusion: = f ) = 0. + 0.008) = 0.3 T = 3 0.x ) 0.3 x ) = 3 0.x ) 0.116x ) 5.6 y=t.8 y=t 1 3. y=f y=t. 1.6 0.8.5 0 0.5 1 1.5.5 3 3.5.5 5 5.5 6 6.5 7 Figure : Linear and quadratic approximations to f = x + x + 5. As is evident in Figure above, the quadratic polynomial T red graph) provides a better approximation than the linear approximation T 1 to the function f when x is close to x 0 =. Moreover, T stays close to f over a wider interval around x 0 =. Definition: The quadratic Taylor polynomial for the function y = f, centered at x 0, fx 0 ), is the function This function has the properties T x 0 ) = fx 0 ) T x 0 ) = f x 0 ) T x 0 ) = f x 0 ) T = fx 0 ) + f x 0 )x x 0 ) + f x 0 ) x x 0 ).

These properties lead to the quadratic approximation formula f T if x x 0. Example. Find the quadratic Taylor polynomial for f = x, centered at x 0 = 5. We need to find f5), f 5) and f 5)... f = x = x 1/ = f = 1 x 1/ and f = 1 x 3/, so f5) = 5 1/ = 5, f 5) = 1 5 1/ = 1 10 and f 5) = 1 5 3/ = 1 500. So... T = fx 0 ) + f x 0 )x x 0 ) + f x 0 ) x x 0 ) = 5 + 1 1 x 5) x 5) 10 1000 To see how good the quadratic approximation is, we can compare it to the linear approximation both graphically and numerically. Graphically: 6 5 3 y=t y=t y=x 1/ 1 0 5 10 15 0 5 30 35 0-1 Numerically: x T 1 T x calculator) x T1 x T 5 5 5 5 0 0.9.899.898979... > 0.001 < 0.00001 6 5.1 5.099 5.099019... > 0.0009 < 0.0000 3.8.796.795831... > 0.00 < 0.00017 7 5. 5.196 5.19615... > 0.003 < 0.00016 0.5.75.7135... > 0.07 < 0.009 30 5.5 5.75 5.775... > 0.0 < 0.003 16.1.019 0.1 0.019 36 6.1 5.979 6 0.1 0.01 5