University:
York University
Course:
MATH 1013 | Applied Calculus I
Academic year:

2022
Views:

326

Pages:

5
Author:

Nick T.

3.1 Exercises a. Number e defined: [\frac{e^h - 1}{h} = 1] [\frac{2.7^h - 1}{h} = 0.99] [\frac{2.8^h - 1}{h} = 1.03] (x^{11} - x + 3) -2x^{-1} (W(u) = 1.1u^{-5}) [-5.4u^{-4}] (f(x) = x^{3/4} + x^{-1/4}) [\frac{3}{4}x^{-1/4} - \frac{1}{4}x^{-5/4}] (f(t) = \frac{1}{t} + \frac{1}{t^4}) [f'(t) = -\frac{1}{t^2} - \frac{4}{t^5}] (y = 2x + \sqrt{x} + x^{4/2}) [y' = 2 + \frac{1}{2\sqrt{x}} + 2x] [y' = 2 + \frac{1}{\sqrt{x}} + x^{1/2}] [y' = -\frac{1}{2}x^{-3/2} + \frac{1}{4}x^{-5/4}] [\frac{1}{\sqrt{x}} + \sqrt[4]{x}] [-\frac{1}{2}x^{-3/2} + \frac{1}{4}x^{-3/4}] [-\frac{1}{2x^{3/2}} + \frac{1}{4x^{3/4}}] [-\frac{1}{2x\sqrt{x}} + \frac{1}{4x\sqrt[4]{x}}] [-\frac{1}{4x\sqrt{x}} - \frac{1}{4x\sqrt[4]{x}} = -\frac{1}{4x}(\sqrt{x} + \sqrt[4]{x})] (x^4 + 8x^3) [4x^3 + 24x^2] (3e^x + \frac{4}{\sqrt[3]{x}} = 4x^{-1/3}) [3e^x - \frac{4}{3}x^{-4/3}] 23. x^2 + x^3 / x To divide the x first 2x + x^2 = 3 + 2x 25. 3y^3/2 + y^5/2 3y^1/2 + y^5/2 27. x^2 + c^4 Why does that * disappear e^2 ln(1.4) 29. A + Bx + Cx^2 e^2 ln(1.4) A + Bx + Cx^2 -1A - B - C -1A - B - C 31. (1 + 16t^2) / (4t)^5 1 + 16t^2 / 64t^5 1/64t^5 + 1/4t 1/64t^5 + 16t^2/64t^5 1/64t^3 + 16t^2/64t^3 1/4t 2 - 3 - 1 -3 - 1 - 4 33. (2w^2 - w + 4) / sqrt(w) 2w^2 - w + 4 / w^1/2 2w^3/2 - w^1/2 + 4w^-1/2 3w^1/2 - 1/2w^-1/2 - 2w^-5/2