Calculators and numerical differentiation

[Albert Chan]

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^(n-1)) * 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

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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