Review 1 Calculus I The exam will apprehend x2.1-6, x3.1-9. The problems will be mainly from the homework and some from the lectures. No calculators or formula sheets will be allowed. A list of adjudicate of problems are given at the end. Chapter 2. Limits 1. De¯nition: limx!a f(x) = L means f(x) is randomly close to L as long as x is close enough (but not equal) to a. 2. Often lim x!a f(x) = f(a). This happens when f(x) is never-ending at x = a: lim x!a f(x) = f(a); 3. If f(x) is not never-ending at x = a, you have to use some tricks: i) Factoring, such(prenominal) as for limx!2 x2¡4 x¡2 = 4; or ii) rationalizing. 4. adept sided limits: limx!1bxc does not exist because limx!1+bxc = 1 while limx!1¡bxc = 0. 5. Trigonometric limits: The basic formula is lim x!0 ungodliness x x = 1; 6. For f(x) = g(x) h(x) , an in¯nite limit occurs at x = a if h(a) = 0 and g(a) 6= 0. Then y = f(x) has a erect asymptote. A limit at in¯nity gives a naiant asymptote. Chapter 3. Derivatives 1. Use the de¯nition to evaluate derivatives of simple functions: x2; p x; 1=x. 2. topaz line at (a; f(a)): y ¡ f(a) = f0(a)(x ¡ a). f0(x) = 0 gives horizontal suntan lines. 3. Use derivatives to ¯nd velocity, acceleration and marginal cost, and to solve falling body problems. 4. Derivative rules.
Sum-di®erence: (f(x) § g(x))0 = f0(x) § g0(x): Product: (f(x)g(x))0 = (f0(x)g(x) + f(x)g0(x)): Quotient: µ f(x) g(x) ¶0 = f0(x)g(x) ¡ f(x)g0(x) g(x)2 : Chain rule: (f ± g)0(x) = f0(g(x))g0(x), or dy dx = dy du du dx , where u = g(x): 5. Derivative formulas. Dx(a) = 0; Dxxr = rxr¡1; Dx(ekx) = kekx (sin x)0 = cos x; (tan x)0 = sec2 x; (sec x)0 = tan x sec x; (cos x)0 = ¡sin x; (cot x)0 = ¡csc2 x; (csc x)0 = ¡cot x csc x; (ln x)0 = 1=x; (arcsin x)0 = 1= p 1 ¡ x2; (arctan x)0 = 1=(1 + x2): 6.Implicit di®erentiation for F(x; y) = 0. Take derivatives of both sides of the equation, treating y as a function of x and apply chain rule. 7. Logarithmic derivation: Note x = eln x. For... If you wishing to get a full essay, order it on our website:
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