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chain rule pdf

The chain rule is the most important and powerful theorem about derivatives. Differentiation: Chain Rule The Chain Rule is used when we want to differentiate a function that may be regarded as a composition of one or more simpler functions. Derivatives - Sum, Power, Product, Quotient, Chain Rules Name_____ ©W X2P0m1q7S xKYu\tfa[ mSTo]fJtTwYa[ryeD OLHLvCr._ ` eAHlblD HrgiIg_hetPsL freeWsWehrTvie]dN.-1-Differentiate each function with respect to x. Chain Rule of Calculus •The chain rule states that derivative of f (g(x)) is f '(g(x)) ⋅g '(x) –It helps us differentiate composite functions •Note that sin(x2)is composite, but sin (x) ⋅x2 is not •sin (x²) is a composite function because it can be constructed as f (g(x)) for f (x)=sin(x)and g(x)=x² –Using the chain rule … Proving the chain rule Given ′ and ′() exist, we want to find . f0(u) = dy du = 3 and g0(x) = du dx = 2). Then differentiate the function. Be able to compare your answer with the direct method of computing the partial derivatives. In other words, we want to compute lim h→0 f(g(x+h))−f(g(x)) h. Many answers: Ex y = (((2x + 1)5 + 2) 6 + 3) 7 dy dx = 7(((2x + 1)5 + 2) 6 + 3) 6 ⋅ 6((2x + 1)5 + 2) 5 ⋅ 5(2x + 1)4 ⋅ 2-2-Create your own worksheets like this … Then, y is a composite function of x; this function is denoted by f g. • In multivariable calculus, you will see bushier trees and more complicated forms of the Chain Rule where you add products of derivatives along paths, Proof of the Chain Rule • Given two functions f and g where g is differentiable at the point x and f is differentiable at the point g(x) = y, we want to compute the derivative of the composite function f(g(x)) at the point x. Then lim →0 = ′ , so is continuous at 0. Problems may contain constants a, b, and c. 1) f (x) = 3x5 2) f (x) = x 3) f (x) = x33 4) f (x) = -2x4 5) f (x) = - 1 4 What if anything can we say about (f g)0(x), the derivative of the composition Note that +− = holds for all . Present your solution just like the solution in Example21.2.1(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 nally show the chain rule being applied to get the answer). Guillaume de l'Hôpital, a French mathematician, also has traces of the y=f(u) u=f(x) y=(2x+4)3 y=u3andu=2x+4 dy du =3u2 du dx =2 dy dx VCE Maths Methods - Chain, Product & Quotient Rules The chain rule 3 • The chain rule is used to di!erentiate a function that has a function within it. If our function f(x) = (g h)(x), where g and h are simpler functions, then the Chain Rule may be stated as f ′(x) = (g h) (x) = (g′ h)(x)h′(x). MATH 200 GOALS Be able to compute partial derivatives with the various versions of the multivariate chain rule. Be able to compute the chain rule based on given values of partial derivatives rather than explicitly defined functions. Let = +− for ≠0 and 0= ′ . 21{1 Use the chain rule to nd the following derivatives. (Section 3.6: Chain Rule) 3.6.2 We can think of y as a function of u, which, in turn, is a function of x. Let’s see this for the single variable case rst. • The chain rule • Questions 2. The chain rule is a simple consequence of the fact that di erentiation produces the linear approximation to a function at a point, and that the derivative is the coe cient appearing in this linear approximation. The Chain Rule Suppose we have two functions, y = f(u) and u = g(x), and we know that y changes at a rate 3 times as fast as u, and that u changes at a rate 2 times as fast as x (ie. 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. Call these functions f and g, respectively. Now let = + − , then += (+ ). 13) Give a function that requires three applications of the chain rule to differentiate. It is especially transparent using o() For a first look at it, let’s approach the last example of last week’s lecture in a different way: Exercise 3.3.11 (revisited and shortened) A stone is dropped into a lake, creating a cir-cular ripple that travels outward at a … Now let = + −, then += ( + ) ( u ) du! Thought to have first originated from the German mathematician Gottfried W. Leibniz ( + ) du... = du dx = 2 ) function that requires three applications of the chain to... ( u ) = dy du = 3 and g0 ( x ) = du dx = 2.! Single variable case rst Gottfried W. Leibniz ′, so is continuous 0... S see this for the single variable case rst f0 ( u ) = du. 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