DOCSLIB.ORG

Gradients, Chain Rule, and Tangent Planes

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Gradients, Chain Rule, and Tangent Planes Properties of the gradient • krf(p)k is maximum rate of increase at p. • rf(p) Recall that for a function f(x, y) to understand lo- points in the direction which gives the p cally how the function is behaving we need to un- maximum rate of increase at . derstand the partial derivatives. For convenience we • -krf(p)k is maximum rate of decrease at p. have lumped these partial derivatives together into a vector which we call the gradient vector and denote • -rf(p) points in the direction which gives the as rf. In particular, the gradient is a vector valued maximum rate of decrease at p. function defined as In particular the gradient encodes information about @f @f how to move to achieve maximum rates of increase rf(x, y) = (x, y), (x, y) . and/or decrease. @x @y On the other hand Du~ f(p) = 0 along level curves If we have a function of three variables, for example or level surfaces. This is because on a level curve or g(x, y, z) then the gradient becomes level surface the function is constant (i.e., unchang- ing). So if ~u is tangent to a level curve we can con- @g @g @g clude that rf(p) is perpendicular to ~u. Or put more rg(x, y, z) = (x, y, z), (x, y, z), (x, y, z) . @x @y @z succinctly: @f rf(p) is perpendicular to level curves/surfaces. If we look at the individual terms we see that @x is indicating how the function is behaving in the x- direction. That is to say that if we start from an initial This gives an easy way to find the normal for tangent point and move in the (positive) x-direction then the planes to a surface, namely given a surface described F( ) = k rF( ) initial rate of change of the function is given by this by p we use p as the normal vector. @f partial derivative. Similarly we have that @y indi- Chain rule cates how the function is behaving in the y-direction. But what if we want to know how the function is The chain rule works for when we have a function behaving in some other direction? inside of a function. For example z = f(x, y), x = x(t) To answer this kind of question we will use direc- and y = y(t). Then z is ultimately a function of t and tional derivatives, so it is natural to ask how does z vary as we vary t, or in other words what is dz . From differentiability f(p + h~u)- f(p) dt Du~ f(p) = lim . we can rearrange the terms to get h 0 h @z @z ∆z = ∆x + ∆y + ERROR Where Du~ f(p) is “the! directional derivative of f in @x @y the direction ~u from the point p = (x0, y0)”. To in- dicate direction we will always use unit vectors. If where ∆z = f(x, y)- f(x0, y0), ∆x = x - x0 and we think of f as elevation and p as indicating our ∆y = y - y0. Also we note that the error term is latitude/longitude position and ~u indicating which very small, much smaller than our other terms and (x y ) direction we want to go, then Du~ f(p) indicates how becomes insignificant near 0, 0 . If we now divide steep it is from our current location if we move in the both sides by ∆t and take a limit as ∆t 0 we have ~u direction indicated by . dz @z dx @z dy We have already mastered doing directional = + . ! dt @x dt @y dt derivatives in the x and y directions (these are our old friends, the partial derivatives). The nice thing In the above expression we have used both “dz” and is that when our function is differentiable finding di- “@z”. We use the “d” when we are treating the func- rectional derivatives in any direction is just as easy. tion as depending on a single variable and the “@” In particular we have the following: when we are treating the function as depending on two or more variables; however they both are asking Du~ f(p) = rf(p) · ~u. to do the same thing, namely take a derivative. The same idea works in more complicated situa- (Usually the hardest thing about these problems is to tions. For instance we could have that z = f(x, y) and remember to make sure that ~u is a unit vector.) that x = x(s, t) and y = y(s, t) so that z depends on Recall that ~u · ~v = k~ukk~vk cos θ, and since ~u is a both s and t. In this case we have unit vector we have D f(p) = krf(p)k cos θ. Since u~ @z @z @x @z @y -1 cos θ 1 this gives bounds on how large the = + 6 6 @s @x @s @y @s directional derivative can be, and in which direction @z @z @x @z @y we achieve maximum and minimum values. In par- = + ticular we have the following: @t @x @t @y @t In general whenever we deal with functions within or in differential form functions (possibly within functions themselves, an dz = f dx + f dy inception of functions if you will) then we can find x y . how the variables change with respect to one another. These types of formula are useful when we want One easy way to keep track is to form a tree showing to approximate the change in output given that we the dependencies among the variables. Then to find know the approximate changes of our input. In par- the derivative of the top variable with respect to one ticular this can be used for error tolerance but we of the leaves we simply add up the product of the can also use this to give approximate values for the partial derivatives of the paths (as discussed in class). function near a point that we can easily evaluate the One special case that we have is when we have im- function. plicit relationships among the variables. For exam- Tangent planes are trying to mimic the function ple, F(x, y) = 0 defines y as a function of x. So F so that it agrees locally with the function in both depends on x and y but the relationship also shows the value of the function and the first order partial that y is a function of x. Taking the derivative of both derivatives of the function. We can also try to find a sides with respect to x using the chain rule we get function that matches the value, the first order partial derivatives and the second order partial derivatives. dy dy -Fx Fx + Fy = 0 or = . Taylor polynomials dx dx F These are done by using the . The y second order Taylor polynomials are shown below Similarly we get the following: (where the function and derivatives are all evaluated at the point (x0, y0)): @z -Fx @z -Fy 1 2 F(x, y, z) = k = = and = . z = f + fx(x - x0) + fy(y - y0) + 2 fxx(x - x0) + @x Fz @y Fz 1 2 fxy(x - x0)(y - y0) + 2 fyy(y - y0) ) Tangent planes and other miscellany Review problems (One to appear on Quiz 6) We have seen tangent planes done in two different 1. On the surface z = x3 - 3xy + y2 a marble is ways. When we did differentiability for a function placed over the point (1, 2). When released the z = f(x, y) we said that a function locally looks like marble initially move in the direction of steepest a plane along with some possible error, the plane we decrease. Find a vector pointing ha, bi pointing got was as follows: in the direction the marble will initially move. @f @f 2. Find the directional derivative for the function z = f(x , y )+ (x , y )(x-x )+ (x , y )(y-y ). 2 0 0 @x 0 0 0 @y 0 0 0 f(x, y, z) = xz - 3xy + 2xyz - 3x + 5y - 17 from the point (2, -6, 3) in the direction of the origin. The second way we have seen these tangent planes is when dealing with F(x, y, z) = k, which we can 3. On the amazing can machine, which has the abil- think of as a level surface. In this case for a point ity to alter the dimensions of cylindrical cans, h = 5 p = (x , y , z ) we can use the properties of gradient the gauges currently read as follows: , 0 0 0 dV = -6π dr = - 1 dh = 1 V h to note that rF(p) will be our normal vector so that dt , dt 2 , and dt , where , , r our tangent plane is are volume, height and radius. But the gauge for r is broken. Find the possible value(s) for r. rF(p) · hx - x0, y - y0, z - z0i = 0. 4. Given the implicitly relationship z + sin z = xy find @2z=@x@y only in terms of z. or if we expand out the above we get the following: 5. Suppose that f(x, y) is differentiable and satisfies 3 2 4 3 @F @F @F f(t -t+1, 2-t ) = t -4t +4t+6. Find fx(1, 1) (p)(x - x0) + (p)(y - y0) + (p)(z - z0) = 0 @x @y @z and fy(1, 1). 6. Let f(x, y) = ex-y1 + sin(2x + 3y). Find an It is important to note that these two definitions are approximate value for ∆z = f(0.2, 0.1)- f(0, 0).
Load more

Recommended publications
  • Anti-Chain Rule Name
    Anti-Chain Rule Name: The most fundamental rule for computing derivatives in calculus is the Chain Rule, from which all other differentiation rules are derived. In words, the Chain Rule states that, to differentiate f(g(x)), differentiate the outside function f and multiply by the derivative of the inside function g, so that d f(g(x)) = f 0(g(x)) · g0(x): dx The Chain Rule makes computation of nearly all derivatives fairly easy. It is tempting to think that there should be an Anti-Chain Rule that makes antidiffer- entiation as easy as the Chain Rule made differentiation. Such a rule would just reverse the process, so instead of differentiating f and multiplying by g0, we would antidifferentiate f and divide by g0. In symbols, Z F (g(x)) f(g(x)) dx = + C: g0(x) Unfortunately, the Anti-Chain Rule does not exist because the formula above is not Z always true. For example, consider (x2 + 1)2 dx. We can compute this integral (correctly) by expanding and applying the power rule: Z Z (x2 + 1)2 dx = x4 + 2x2 + 1 dx x5 2x3 = + + x + C 5 3 However, we could also think about our integrand above as f(g(x)) with f(x) = x2 and g(x) = x2 + 1. Then we have F (x) = x3=3 and g0(x) = 2x, so the Anti-Chain Rule would predict that Z F (g(x)) (x2 + 1)2 dx = + C g0(x) (x2 + 1)3 = + C 3 · 2x x6 + 3x4 + 3x2 + 1 = + C 6x x5 x3 x 1 = + + + + C 6 2 2 6x Since these answers are different, we conclude that the Anti-Chain Rule has failed us.
    [Show full text]
  • 9.2. Undoing the Chain Rule
    < Previous Section Home Next Section > 9.2. Undoing the Chain Rule This chapter focuses on finding accumulation functions in closed form when we know its rate of change function. We’ve seen that 1) an accumulation function in closed form is advantageous for quickly and easily generating many values of the accumulating quantity, and 2) the key to finding accumulation functions in closed form is the Fundamental Theorem of Calculus. The FTC says that an accumulation function is the antiderivative of the given rate of change function (because rate of change is the derivative of accumulation). In Chapter 6, we developed many derivative rules for quickly finding closed form rate of change functions from closed form accumulation functions. We also made a connection between the form of a rate of change function and the form of the accumulation function from which it was derived. Rather than inventing a whole new set of techniques for finding antiderivatives, our mindset as much as possible will be to use our derivatives rules in reverse to find antiderivatives. We worked hard to develop the derivative rules, so let’s keep using them to find antiderivatives! Thus, whenever you have a rate of change function f and are charged with finding its antiderivative, you should frame the task with the question “What function has f as its derivative?” For simple rate of change functions, this is easy, as long as you know your derivative rules well enough to apply them in reverse. For example, given a rate of change function …. … 2x, what function has 2x as its derivative? i.e.
    [Show full text]
  • Calculus Terminology
    AP Calculus BC Calculus Terminology Absolute Convergence Asymptote Continued Sum Absolute Maximum Average Rate of Change Continuous Function Absolute Minimum Average Value of a Function Continuously Differentiable Function Absolutely Convergent Axis of Rotation Converge Acceleration Boundary Value Problem Converge Absolutely Alternating Series Bounded Function Converge Conditionally Alternating Series Remainder Bounded Sequence Convergence Tests Alternating Series Test Bounds of Integration Convergent Sequence Analytic Methods Calculus Convergent Series Annulus Cartesian Form Critical Number Antiderivative of a Function Cavalieri’s Principle Critical Point Approximation by Differentials Center of Mass Formula Critical Value Arc Length of a Curve Centroid Curly d Area below a Curve Chain Rule Curve Area between Curves Comparison Test Curve Sketching Area of an Ellipse Concave Cusp Area of a Parabolic Segment Concave Down Cylindrical Shell Method Area under a Curve Concave Up Decreasing Function Area Using Parametric Equations Conditional Convergence Definite Integral Area Using Polar Coordinates Constant Term Definite Integral Rules Degenerate Divergent Series Function Operations Del Operator e Fundamental Theorem of Calculus Deleted Neighborhood Ellipsoid GLB Derivative End Behavior Global Maximum Derivative of a Power Series Essential Discontinuity Global Minimum Derivative Rules Explicit Differentiation Golden Spiral Difference Quotient Explicit Function Graphic Methods Differentiable Exponential Decay Greatest Lower Bound Differential
    [Show full text]
  • A General Chain Rule for Distributional Derivatives
    proceedings of the american mathematical society Volume 108, Number 3, March 1990 A GENERAL CHAIN RULE FOR DISTRIBUTIONAL DERIVATIVES L. AMBROSIO AND G. DAL MASO (Communicated by Barbara L. Keyfitz) Abstract. We prove a general chain rule for the distribution derivatives of the composite function v(x) = f(u(x)), where u: R" —>Rm has bounded variation and /: Rm —>R* is Lipschitz continuous. Introduction The aim of the present paper is to prove a chain rule for the distributional derivative of the composite function v(x) = f(u(x)), where u: Q —>Rm has bounded variation in the open set ilcR" and /: Rw —>R is uniformly Lip- schitz continuous. Under these hypotheses it is easy to prove that the function v has locally bounded variation in Q, hence its distributional derivative Dv is a Radon measure in Q with values in the vector space Jz? m of all linear maps from R" to Rm . The problem is to give an explicit formula for Dv in terms of the gradient Vf of / and of the distributional derivative Du . To illustrate our formula, we begin with the simpler case, studied by A. I. Vol pert, where / is continuously differentiable. Let us denote by Su the set of all jump points of u, defined as the set of all xef! where the approximate limit u(x) does not exist at x. Then the following identities hold in the sense of measures (see [19] and [20]): (0.1) Dv = Vf(ü)-Du onß\SH, and (0.2) Dv = (f(u+)-f(u-))®vu-rn_x onSu, where vu denotes the measure theoretical unit normal to Su, u+ , u~ are the approximate limits of u from both sides of Su, and %?n_x denotes the (n - l)-dimensional Hausdorff measure.
    [Show full text]
  • Math 212-Lecture 8 13.7: the Multivariable Chain Rule
    Math 212-Lecture 8 13.7: The multivariable chain rule The chain rule with one independent variable w = f(x; y). If the particle is moving along a curve x = x(t); y = y(t), then the values that the particle feels is w = f(x(t); y(t)). Then, w = w(t) is a function of t. x; y are intermediate variables and t is the independent variable. The chain rule says: If both fx and fy are continuous, then dw = f x0(t) + f y0(t): dt x y Intuitively, the total change rate is the superposition of contributions in both directions. Comment: You cannot do as in one single variable calculus like @w dx @w dy dw dw + = + ; @x dt @y dt dt dt @w by canceling the two dx's. @w in @x is only the `partial' change due to the dw change of x, which is different from the dw in dt since the latter is the total change. The proof is to use the increment ∆w = wx∆x + wy∆y + small. Read Page 903. Example: Suppose w = sin(uv) where u = 2s and v = s2. Compute dw ds at s = 1. Solution. By chain rule dw @w du @w dv = + = cos(uv)v ∗ 2 + cos(uv)u ∗ 2s: ds @u ds @v ds At s = 1, u = 2; v = 1. Hence, we have cos(2) ∗ 2 + cos(2) ∗ 2 ∗ 2 = 6 cos(2): We can actually check that w = sin(uv) = sin(2s3). Hence, w0(s) = 3 2 cos(2s ) ∗ 6s . At s = 1, this is 6 cos(2).
    [Show full text]
  • Chapter 4 Differentiation in the Study of Calculus of Functions of One Variable, the Notions of Continuity, Differentiability and Integrability Play a Central Role
    Chapter 4 Differentiation In the study of calculus of functions of one variable, the notions of continuity, differentiability and integrability play a central role. The previous chapter was devoted to continuity and its consequences and the next chapter will focus on integrability. In this chapter we will define the derivative of a function of one variable and discuss several important consequences of differentiability. For example, we will show that differentiability implies continuity. We will use the definition of derivative to derive a few differentiation formulas but we assume the formulas for differentiating the most common elementary functions are known from a previous course. Similarly, we assume that the rules for differentiating are already known although the chain rule and some of its corollaries are proved in the solved problems. We shall not emphasize the various geometrical and physical applications of the derivative but will concentrate instead on the mathematical aspects of differentiation. In particular, we present several forms of the mean value theorem for derivatives, including the Cauchy mean value theorem which leads to L’Hôpital’s rule. This latter result is useful in evaluating so called indeterminate limits of various kinds. Finally we will discuss the representation of a function by Taylor polynomials. The Derivative Let fx denote a real valued function with domain D containing an L ? neighborhood of a point x0 5 D; i.e. x0 is an interior point of D since there is an L ; 0 such that NLx0 D. Then for any h such that 0 9 |h| 9 L, we can define the difference quotient for f near x0, fx + h ? fx D fx : 0 0 4.1 h 0 h It is well known from elementary calculus (and easy to see from a sketch of the graph of f near x0 ) that Dhfx0 represents the slope of a secant line through the points x0,fx0 and x0 + h,fx0 + h.
    [Show full text]
  • The Chain Rule in Partial Differentiation
    THE CHAIN RULE IN PARTIAL DIFFERENTIATION 1 Simple chain rule If u = u(x, y) and the two independent variables x and y are each a function of just one other variable t so that x = x(t) and y = y(t), then to find du/dt we write down the differential of u ∂u ∂u δu = δx + δy + .... (1) ∂x ∂y Then taking limits δx 0, δy 0 and δt 0 in the usual way we have → → → du ∂u dx ∂u dy = + . (2) dt ∂x dt ∂y dt Note we only need straight ‘d’s’ in dx/dt and dy/dt because x and y are function of one variable t whereas u is a function of both x and y. 2 Chain rule for two sets of independent variables If u = u(x, y) and the two independent variables x, y are each a function of two new independent variables s, t then we want relations between their partial derivatives. 1. When u = u(x, y), for guidance in working out the chain rule, write down the differential ∂u ∂u δu = δx + δy + ... (3) ∂x ∂y then when x = x(s, t) and y = y(s, t) (which are known functions of s and t), the chain rule for us and ut in terms of ux and uy is ∂u ∂u ∂x ∂u ∂y = + (4) ∂s ∂x ∂s ∂y ∂s ∂u ∂u ∂x ∂u ∂y = + . (5) ∂t ∂x ∂t ∂y ∂t 2. Conversely, when u = u(s, t), for guidance in working out the chain rule write down the differential ∂u ∂u δu = δs + δt + ..
    [Show full text]
  • Failure of the Chain Rule for the Divergence of Bounded Vector Fields
    FAILURE OF THE CHAIN RULE FOR THE DIVERGENCE OF BOUNDED VECTOR FIELDS GIANLUCA CRIPPA, NIKOLAY GUSEV, STEFANO SPIRITO, AND EMIL WIEDEMANN ABSTRACT. We provide a vast class of counterexamples to the chain rule for the divergence of bounded vector fields in three space dimensions. Our convex integration approach allows us to produce renormalization defects of various kinds, which in a sense quantify the breakdown of the chain rule. For instance, we can construct defects which are absolutely continuous with respect to the Lebesgue measure, or defects which are not even measures. MSC (2010): 35F05 (PRIMARY); 35A02, 35Q35 KEYWORDS: Chain Rule, Convex Integration, Transport and Continuity Equations, Renormaliza- tion 1. INTRODUCTION In this paper we consider the classical problem of the chain rule for the divergence of a bounded vector field. Specifically, the problem can be stated in the following way: d d Let W ⊂ R be a domain with Lipschitz boundary. Given a bounded vector field v : W ! R tangent to the boundary and a bounded scalar function r : W ! R, one asks whether it is possible to express the quantity div(b(r)v), where b is a smooth scalar function, only in terms of b, r and the quantities m = divv and l = div(rv). Indeed, formally we should have that div(b(r)v) = (b(r) − rb 0(r))m + b 0(r)l: (1.1) However, the extension of (1.1) to a nonsmooth setting is far from trivial. The chain rule problem is particularly important in view of its applications to the uniqueness and compactness of transport and continuity equations, whose analysis is nowadays a fundamental tool in the study of various equations arising in mathematical physics.
    [Show full text]
  • Chain Rule & Implicit Differentiation
    INTRODUCTION The chain rule and implicit differentiation are techniques used to easily differentiate otherwise difficult equations. Both use the rules for derivatives by applying them in slightly different ways to differentiate the complex equations without much hassle. In this presentation, both the chain rule and implicit differentiation will be shown with applications to real world problems. DEFINITION Chain Rule Implicit Differentiation A way to differentiate A way to take the derivative functions within of a term with respect to functions. another variable without having to isolate either variable. HISTORY The Chain Rule is thought to have first originated from the German mathematician Gottfried W. Leibniz. Although the memoir it was first found in contained various mistakes, it is apparent that he used chain rule in order to differentiate a polynomial inside of a square root. Guillaume de l'Hôpital, a French mathematician, also has traces of the chain rule in his Analyse des infiniment petits. HISTORY Implicit differentiation was developed by the famed physicist and mathematician Isaac Newton. He applied it to various physics problems he came across. In addition, the German mathematician Gottfried W. Leibniz also developed the technique independently of Newton around the same time period. EXAMPLE 1: CHAIN RULE Find the derivative of the following using chain rule y=(x2+5x3-16)37 EXAMPLE 1: CHAIN RULE Step 1: Define inner and outer functions y=(x2+5x3-16)37 EXAMPLE 1: CHAIN RULE Step 2: Differentiate outer function via power rule y’=37(x2+5x3-16)36 EXAMPLE 1: CHAIN RULE Step 3: Differentiate inner function and multiply by the answer from the previous step y’=37(x2+5x3-16)36(2x+15x2) EXAMPLE 2: CHAIN RULE A biologist must use the chain rule to determine how fast a given bacteria population is growing at a given point in time t days later.
    [Show full text]
  • Lecture 10 : Chain Rule (Please Review Composing Functions Under Algebra/Precalculus Review on the Class Webpage.) Here We Apply the Derivative to Composite Functions
    Lecture 10 : Chain Rule (Please review Composing Functions under Algebra/Precalculus Review on the class webpage.) Here we apply the derivative to composite functions. We get the following rule of differentiation: The Chain Rule : If g is a differentiable function at x and f is differentiable at g(x), then the composite function F = f ◦ g defined by F (x) = f(g(x)) is differentiable at x and F 0 is given by the product F 0(x) = f 0(g(x))g0(x): In Leibniz notation If y = f(u) and u = g(x) are both differentiable functions, then dy dy du = : dx du dx It is not difficult to se why this is true, if we examine the average change in the value of F (x) that results from a small change in the value of x: F (x + h) − F (x) f(g(x + h)) − f(g(x)) f(g(x + h)) − f(g(x)) g(x + h) − g(x) = = · h h g(x + h) − g(x) h or if we let u = g(x) and y = F (x) = f(u), then ∆y ∆y ∆u = · ∆x ∆u ∆x if g(x + h) − g(x) = ∆u 6= 0. When we take the limit as h ! 0 or ∆x ! 0, we get F 0(x) = f 0(g(x))g0(x) or dy dy du = : dx du dx Example Find the derivative of F (x) = sin(2x + 1). Step 1: Write F (x) as F (x) = f(g(x)) or y = F (x) = f(u), where u = g(x).
    [Show full text]
  • Today in Physics 217: Vector Derivatives
    Today in Physics 217: vector derivatives First derivatives: • Gradient (—) • Divergence (—◊) •Curl (—¥) Second derivatives: the Laplacian (—2) and its relatives Vector-derivative identities: relatives of the chain rule, product rule, etc. Image by Eric Carlen, School of Mathematics, Georgia Institute of 22 vxyxy, =−++ x yˆˆ x y Technology ()()() 6 September 2002 Physics 217, Fall 2002 1 Differential vector calculus df/dx provides us with information on how quickly a function of one variable, f(x), changes. For instance, when the argument changes by an infinitesimal amount, from x to x+dx, f changes by df, given by df df= dx dx In three dimensions, the function f will in general be a function of x, y, and z: f(x, y, z). The change in f is equal to ∂∂∂fff df=++ dx dy dz ∂∂∂xyz ∂∂∂fff =++⋅++ˆˆˆˆˆˆ xyzxyz ()()dx() dy() dz ∂∂∂xyz ≡⋅∇fdl 6 September 2002 Physics 217, Fall 2002 2 Differential vector calculus (continued) The vector derivative operator — (“del”) ∂∂∂ ∇ =++xyzˆˆˆ ∂∂∂xyz produces a vector when it operates on scalar function f(x,y,z). — is a vector, as we can see from its behavior under coordinate rotations: I ′ ()∇∇ff=⋅R but its magnitude is not a number: it is an operator. 6 September 2002 Physics 217, Fall 2002 3 Differential vector calculus (continued) There are three kinds of vector derivatives, corresponding to the three kinds of multiplication possible with vectors: Gradient, the analogue of multiplication by a scalar. ∇f Divergence, like the scalar (dot) product. ∇⋅v Curl, which corresponds to the vector (cross) product. ∇×v 6 September 2002 Physics 217, Fall 2002 4 Gradient The result of applying the vector derivative operator on a scalar function f is called the gradient of f: ∂∂∂fff ∇f =++xyzˆˆˆ ∂∂∂xyz The direction of the gradient points in the direction of maximum increase of f (i.e.
    [Show full text]
  • 1. Antiderivatives for Exponential Functions Recall That for F(X) = Ec⋅X, F ′(X) = C ⋅ Ec⋅X (For Any Constant C)
    1. Antiderivatives for exponential functions Recall that for f(x) = ec⋅x, f ′(x) = c ⋅ ec⋅x (for any constant c). That is, ex is its own derivative. So it makes sense that it is its own antiderivative as well! Theorem 1.1 (Antiderivatives of exponential functions). Let f(x) = ec⋅x for some 1 constant c. Then F x ec⋅c D, for any constant D, is an antiderivative of ( ) = c + f(x). 1 c⋅x ′ 1 c⋅x c⋅x Proof. Consider F (x) = c e +D. Then by the chain rule, F (x) = c⋅ c e +0 = e . So F (x) is an antiderivative of f(x). Of course, the theorem does not work for c = 0, but then we would have that f(x) = e0 = 1, which is constant. By the power rule, an antiderivative would be F (x) = x + C for some constant C. 1 2. Antiderivative for f(x) = x We have the power rule for antiderivatives, but it does not work for f(x) = x−1. 1 However, we know that the derivative of ln(x) is x . So it makes sense that the 1 antiderivative of x should be ln(x). Unfortunately, it is not. But it is close. 1 1 Theorem 2.1 (Antiderivative of f(x) = x ). Let f(x) = x . Then the antiderivatives of f(x) are of the form F (x) = ln(SxS) + C. Proof. Notice that ln(x) for x > 0 F (x) = ln(SxS) = . ln(−x) for x < 0 ′ 1 For x > 0, we have [ln(x)] = x .
    [Show full text]