Question

What is the difference between the approximation and the actual value of f′(0.5)?

Answer 1

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

$f'(0.5)=\frac{(2-1)}{(1-0)}\\\\f'(0.5)=1---(2)$

Given function is

$f(x)=2^{x^{3}}$

Differentiating w.r.to x

$f'(x)=ln(2)2^{x^{3}}\frac{d}{dx}(x^{3})\\\\f'(x)=ln(2)2^{x^{3}}(3x^{2})\\\\f'(x)=3x^{2}2^{x^{3}}ln(2)\\at \quad x=0.5\\\\f'(0.5)=0.5669---(2)$

Difference between the approximation and the actual value of f′(0.5) is 0.4330