Answer:
Difference between the approximation and the actual value of f′(0.5) is 0.4330
Step-by-step explanation:
As complete question is not give, so considering the most relevant question to given statement in Fig below.
from table
x 0 1
f(x) 1 2
f'(0.5)=?
from numerical differentiation
[tex]f'(0.5)=\frac{(2-1)}{(1-0)}\\\\f'(0.5)=1---(2)[/tex]
Given function is
[tex]f(x)=2^{x^{3}}[/tex]
Differentiating w.r.to x
[tex]f'(x)=ln(2)2^{x^{3}}\frac{d}{dx}(x^{3})\\\\f'(x)=ln(2)2^{x^{3}}(3x^{2})\\\\f'(x)=3x^{2}2^{x^{3}}ln(2)\\at \quad x=0.5\\\\f'(0.5)=0.5669---(2)[/tex]
Difference between the approximation and the actual value of f′(0.5) is 0.4330
