Calculators and numerical differentiation

11032020, 10:14 PM
(This post was last modified: 11042020 12:48 AM by Albert Chan.)
Post: #9




RE: Calculators and numerical differentiation
Slightly off topics, for f(x) = x*g(x), getting f'(0) is easier taking limit directly.
\(f(x) = x·g(x) = x·\sqrt[3]{x^2+x}\) \(f'(0) = \displaystyle{\lim_{h \to 0}} {f(h)f(0)\over h} = \displaystyle{\lim_{h \to 0}}\; g(h) = g(0) = 0\) Getting derivative take more work, and easier to make mistakes. f' = (x*g)' = g + x*g' g has the form z^n, where z = x²+x, n = 1/3 g' = (n*z^(n1)) * z' = (n*g) * (z'/z) = g/3 * (2x+1)/(x²+x) f' = g + g/3 * (2x+1)/(x+1) = g * (5x+4)/(3x+3) f'(0) = g(0) * 4/3 = 0  Another way, by shape of the curve. \(f(x) = x·\sqrt[3]{x^2+x} = \large\frac{\sqrt[3]{(x^2+x)^4}}{x+1}\) f(ε) > 0, f(ε) > 0, f(0) = 0 → f(0) is local minimum → f'(0) = 0 

« Next Oldest  Next Newest »

Messages In This Thread 
Calculators and numerical differentiation  robve  10302020, 09:57 PM
RE: Calculators and numerical differentiation  Paul Dale  10302020, 11:41 PM
RE: Calculators and numerical differentiation  Albert Chan  10312020, 01:20 AM
RE: Calculators and numerical differentiation  Wes Loewer  11012020, 05:39 AM
RE: Calculators and numerical differentiation  Albert Chan  11012020, 05:39 PM
RE: Calculators and numerical differentiation  Albert Chan  11012020, 11:43 PM
RE: Calculators and numerical differentiation  Wes Loewer  11032020, 06:09 PM
RE: Calculators and numerical differentiation  Albert Chan  11032020 10:14 PM
RE: Calculators and numerical differentiation  Wes Loewer  11042020, 04:14 PM
RE: Calculators and numerical differentiation  CMarangon  11032020, 06:55 PM
RE: Calculators and numerical differentiation  Wes Loewer  11042020, 04:04 PM

User(s) browsing this thread: 1 Guest(s)