DOCSLIB.ORG

21. Chain Rule Chain Rule

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
21. Chain Rule Chain Rule 21. Chain rule Chain rule Statement 21.1. Statement Examples The power rule says that d [xn] = nxn−1: dx This rule is valid for any power n, but not for any base other than the simple input variable x. For instance, d (2x + 5)3 6= 3(2x + 5)2: dx The function h(x) = (2x + 5)3 is built up of the two simpler functions g(x) = 2x + 5 and Table of Contents f(x) = x3: h(x) = (2x + 5)3 = (g(x))3 = f(g(x)): JJ II Technically speaking, h is the composition of f and g. The next rule expresses the derivative of such a function in terms of the derivatives of its components. J I Chain rule. For functions f and g Page1 of8 d [f(g(x))] = f 0(g(x)) · g0(x): dx Back Print Version In the composition f(g(x)), we call f the outside function and g the inside function. With this terminology, the rule says that the derivative of the composition of two functions is Home Page the derivative of the outside function evaluated at the inside function times the derivative of the inside function. 21.2. Examples Chain rule d Statement 21.2.1 Example Find the derivative (2x + 5)3. dx Examples Solution We begin by viewing (2x + 5)3 as a composition of functions and identifying the outside function f and the inside function g. The outside function is the last thing you do when computing the expression for a given input x. Here, the outside function is the cubing function: (2x + 5)3 = f(g(x)); where f(x) = x3 and g(x) = 2x + 5: Next, we do the computations required for the chain rule formula: Table of Contents f(x) = x3 g(x) = 2x + 5 f 0(x) = 3x2 g0(x) = 2 JJ II f 0(g(x)) = 3(2x + 5)2 J I Finally, we use the formula: d Page2 of8 [f(g(x))] = f 0(g(x)) · g0(x) dx ### Back d (2x + 5)3 = 3(2x + 5)2 · 2 dx Print Version Home Page d 21.2.2 Example Find the derivative sin(x5). dx Solution Here, the outside function is the sine function: Chain rule sin(x5) = f(g(x)); where f(x) = sin x and g(x) = x5: Statement Examples So f(x) = sin x g(x) = x5 f 0(x) = cos x g0(x) = 5x4 f 0(g(x)) = cos(x5) giving d [f(g(x))] = f 0(g(x)) · g0(x) dx Table of Contents ### d sin(x5) = cos(x5) · 5x4 dx JJ II J I d 21.2.3 Example Find the derivative sin5 x. dx Page3 of8 Solution Recalling that sin5 x means (sin x)5, we see that the outside function is the one Back that raises an input to the fifth power: sin5 x = f(g(x)); where f(x) = x5 and g(x) = sin x: Print Version In order to reduce the number of steps, we go immediately to the chain rule formula and Home Page do the intermediate computations mentally as required: d [f(g(x))] = f 0(g(x)) · g0(x) Chain rule dx ### Statement d sin5 x = 5(sin x)4 · cos x Examples dx d h 2 i 21.2.4 Example Find the derivative 5x −4x+3 . dx Solution Here, the outside function is the exponential function with base 5: 2 5x −4x+3 = f(g(x)); where f(x) = 5x and g(x) = x2 − 4x + 3: Table of Contents Trimming the number of steps a bit more, we omit the formula for the chain rule and just think \Derivative of outside function, evaluated at inside function, times derivative of JJ II inside function": d h 2 i 2 5x −4x+3 = 5x −4x+3 ln 5 · (2x − 4): J I dx Page4 of8 d p 21.2.5 Example Find the derivative 5ex + 4x3 : dx Back Solution The outside function is the square root function: Print Version h i d p x 3 1 x 3 −1=2 x 2 5e + 4x = 2 (5e + 4x ) (5e + 12x ): dx Home Page d h 4 i 21.2.6 Example Find the derivative cos ex . Chain rule dx Statement Solution The outside function is the cosine function: Examples d h 4 i 4 d h 4 i cos ex = − sin ex · ex dx dx 4 4 = − sin ex · ex (4x3): The second step required another use of the chain rule (with outside function the exponen- tial function). ex 21.2.7 Example Find the derivative of f(x) = ee . Table of Contents Solution The chain rule is used twice, each time with outside function the exponential function: JJ II d h ex i f 0(x) = ee J I dx ex d h x i = ee · ee dx Page5 of8 ex x = ee · ee · ex: Back Print Version x3 + 2 21.2.8 Example Find the derivative of f(x) = sin p : 5x − 74x Home Page Solution We have Chain rule d x3 + 2 f 0(x) = sin p Statement dx 5x − 74x Examples x3 + 2 d x3 + 2 = cos p · p 5x − 74x dx 5x − 74x p p 4x d 3 3 d 4x x3 + 2 5x − 7 x + 2 − (x + 2) 5x − 7 = cos p · dx p dx 5x − 74x ( 5x − 74x)2 x3 + 2 = cos p 5x − 74x p 5x − 74x(3x2) − (x3 + 2) 1 (5x − 74x)−1=2(5 − 74x(ln 7)(4)) Table of Contents · 2 : 5x − 74x JJ II J I Page6 of8 Back Print Version Home Page 21 { Exercises Chain rule Statement Examples 21 { 1 Use the chain rule to find the following derivatives. Present your solution just like the solution in Example 21.2.1 (i.e., write the given function as a composition of two functions f and g, compute the quantities required on the right-hand side of the chain rule formula, and finally show the chain rule being applied to get the answer). d (a) (2x4 − 5x)7. dx d Table of Contents (b) [cos(8t − 3)]. dt JJ II 21 { 2 Find the derivatives of the following functions. (You may omit details and proceed as in Example 21.2.5.) J I p (a) f(x) = 3 8 − x3. Page7 of8 (b) f(t) = esin 2t. Back 3 21 { 3 Find the derivative of f(x) = 2x +4x cos(3x). Print Version Home Page e5t 21 { 4 Find the derivative of f(t) = cos . t − t8 Chain rule Statement Examples Table of Contents JJ II J I Page8 of8 Back Print Version Home Page.
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]
  • Infinitesimal Calculus
    Infinitesimal Calculus Δy ΔxΔy and “cannot stand” Δx • Derivative of the sum/difference of two functions (x + Δx) ± (y + Δy) = (x + y) + Δx + Δy ∴ we have a change of Δx + Δy. • Derivative of the product of two functions (x + Δx)(y + Δy) = xy + Δxy + xΔy + ΔxΔy ∴ we have a change of Δxy + xΔy. • Derivative of the product of three functions (x + Δx)(y + Δy)(z + Δz) = xyz + Δxyz + xΔyz + xyΔz + xΔyΔz + ΔxΔyz + xΔyΔz + ΔxΔyΔ ∴ we have a change of Δxyz + xΔyz + xyΔz. • Derivative of the quotient of three functions x Let u = . Then by the product rule above, yu = x yields y uΔy + yΔu = Δx. Substituting for u its value, we have xΔy Δxy − xΔy + yΔu = Δx. Finding the value of Δu , we have y y2 • Derivative of a power function (and the “chain rule”) Let y = x m . ∴ y = x ⋅ x ⋅ x ⋅...⋅ x (m times). By a generalization of the product rule, Δy = (xm−1Δx)(x m−1Δx)(x m−1Δx)...⋅ (xm −1Δx) m times. ∴ we have Δy = mx m−1Δx. • Derivative of the logarithmic function Let y = xn , n being constant. Then log y = nlog x. Differentiating y = xn , we have dy dy dy y y dy = nxn−1dx, or n = = = , since xn−1 = . Again, whatever n−1 y dx x dx dx x x x the differentials of log x and log y are, we have d(log y) = n ⋅ d(log x), or d(log y) n = . Placing these values of n equal to each other, we obtain d(log x) dy d(log y) y dy = .
    [Show full text]
  • Calculus Formulas and Theorems
    Formulas and Theorems for Reference I. Tbigonometric Formulas l. sin2d+c,cis2d:1 sec2d l*cot20:<:sc:20 +.I sin(-d) : -sitt0 t,rs(-//) = t r1sl/ : -tallH 7. sin(A* B) :sitrAcosB*silBcosA 8. : siri A cos B - siu B <:os,;l 9. cos(A+ B) - cos,4cos B - siuA siriB 10. cos(A- B) : cosA cosB + silrA sirrB 11. 2 sirrd t:osd 12. <'os20- coS2(i - siu20 : 2<'os2o - I - 1 - 2sin20 I 13. tan d : <.rft0 (:ost/ I 14. <:ol0 : sirrd tattH 1 15. (:OS I/ 1 16. cscd - ri" 6i /F tl r(. cos[I ^ -el : sitt d \l 18. -01 : COSA 215 216 Formulas and Theorems II. Differentiation Formulas !(r") - trr:"-1 Q,:I' ]tra-fg'+gf' gJ'-,f g' - * (i) ,l' ,I - (tt(.r))9'(.,') ,i;.[tyt.rt) l'' d, \ (sttt rrJ .* ('oqI' .7, tJ, \ . ./ stll lr dr. l('os J { 1a,,,t,:r) - .,' o.t "11'2 1(<,ot.r') - (,.(,2.r' Q:T rl , (sc'c:.r'J: sPl'.r tall 11 ,7, d, - (<:s<t.r,; - (ls(].]'(rot;.r fr("'),t -.'' ,1 - fr(u") o,'ltrc ,l ,, 1 ' tlll ri - (l.t' .f d,^ --: I -iAl'CSllLl'l t!.r' J1 - rz 1(Arcsi' r) : oT Il12 Formulas and Theorems 2I7 III. Integration Formulas 1. ,f "or:artC 2. [\0,-trrlrl *(' .t "r 3. [,' ,t.,: r^x| (' ,I 4. In' a,,: lL , ,' .l 111Q 5. In., a.r: .rhr.r' .r r (' ,l f 6. sirr.r d.r' - ( os.r'-t C ./ 7. /.,,.r' dr : sitr.i'| (' .t 8. tl:r:hr sec,rl+ C or ln Jccrsrl+ C ,f'r^rr f 9.
    [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]
  • Integration by Parts
    3 Integration By Parts Formula ∫∫udv = uv − vdu I. Guidelines for Selecting u and dv: (There are always exceptions, but these are generally helpful.) “L-I-A-T-E” Choose ‘u’ to be the function that comes first in this list: L: Logrithmic Function I: Inverse Trig Function A: Algebraic Function T: Trig Function E: Exponential Function Example A: ∫ x3 ln x dx *Since lnx is a logarithmic function and x3 is an algebraic function, let: u = lnx (L comes before A in LIATE) dv = x3 dx 1 du = dx x x 4 v = x 3dx = ∫ 4 ∫∫x3 ln xdx = uv − vdu x 4 x 4 1 = (ln x) − dx 4 ∫ 4 x x 4 1 = (ln x) − x 3dx 4 4 ∫ x 4 1 x 4 = (ln x) − + C 4 4 4 x 4 x 4 = (ln x) − + C ANSWER 4 16 www.rit.edu/asc Page 1 of 7 Example B: ∫sin x ln(cos x) dx u = ln(cosx) (Logarithmic Function) dv = sinx dx (Trig Function [L comes before T in LIATE]) 1 du = (−sin x) dx = − tan x dx cos x v = ∫sin x dx = − cos x ∫sin x ln(cos x) dx = uv − ∫ vdu = (ln(cos x))(−cos x) − ∫ (−cos x)(− tan x)dx sin x = −cos x ln(cos x) − (cos x) dx ∫ cos x = −cos x ln(cos x) − ∫sin x dx = −cos x ln(cos x) + cos x + C ANSWER Example C: ∫sin −1 x dx *At first it appears that integration by parts does not apply, but let: u = sin −1 x (Inverse Trig Function) dv = 1 dx (Algebraic Function) 1 du = dx 1− x 2 v = ∫1dx = x ∫∫sin −1 x dx = uv − vdu 1 = (sin −1 x)(x) − x dx ∫ 2 1− x ⎛ 1 ⎞ = x sin −1 x − ⎜− ⎟ (1− x 2 ) −1/ 2 (−2x) dx ⎝ 2 ⎠∫ 1 = x sin −1 x + (1− x 2 )1/ 2 (2) + C 2 = x sin −1 x + 1− x 2 + C ANSWER www.rit.edu/asc Page 2 of 7 II.
    [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]
  • Derivative, Gradient, and Lagrange Multipliers
    EE263 Prof. S. Boyd Derivative, Gradient, and Lagrange Multipliers Derivative Suppose f : Rn → Rm is differentiable. Its derivative or Jacobian at a point x ∈ Rn is × denoted Df(x) ∈ Rm n, defined as ∂fi (Df(x))ij = , i =1, . , m, j =1,...,n. ∂x j x The first order Taylor expansion of f at (or near) x is given by fˆ(y)= f(x)+ Df(x)(y − x). When y − x is small, f(y) − fˆ(y) is very small. This is called the linearization of f at (or near) x. As an example, consider n = 3, m = 2, with 2 x1 − x f(x)= 2 . x1x3 Its derivative at the point x is 1 −2x2 0 Df(x)= , x3 0 x1 and its first order Taylor expansion near x = (1, 0, −1) is given by 1 1 1 0 0 fˆ(y)= + y − 0 . −1 −1 0 1 −1 Gradient For f : Rn → R, the gradient at x ∈ Rn is denoted ∇f(x) ∈ Rn, and it is defined as ∇f(x)= Df(x)T , the transpose of the derivative. In terms of partial derivatives, we have ∂f ∇f(x)i = , i =1,...,n. ∂x i x The first order Taylor expansion of f at x is given by fˆ(x)= f(x)+ ∇f(x)T (y − x). 1 Gradient of affine and quadratic functions You can check the formulas below by working out the partial derivatives. For f affine, i.e., f(x)= aT x + b, we have ∇f(x)= a (independent of x). × For f a quadratic form, i.e., f(x)= xT Px with P ∈ Rn n, we have ∇f(x)=(P + P T )x.
    [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]