suppose f(x) = exp(x), and we want to find f’(x) at xi =𝟏. what is the true value of f’(xi ) at xi =𝟏? (use basic calculus) in matlab

To find the derivative of f(x) = exp(x) at x = 1 using basic calculus, we can use the definition of the derivative:

f'(xi) = lim(h->0) [f(xi + h) - f(xi)] / h

In this case, we want to find f'(1).

Substituting xi = 1 into the definition, we have:

f'(1) = lim(h->0) [f(1 + h) - f(1)] / h

Now let's simplify this expression:

f(1 + h) = exp(1 + h) f(1) = exp(1)

f'(1) = lim(h->0) [exp(1 + h) - exp(1)] / h

Now we can use MATLAB to approximate this limit by choosing a small value for h:

h = 1e-6; % small value for h f_prime = (exp(1 + h) - exp(1)) / h;

The value of f'(xi) at xi = 1 is approximately equal to:

f'(1) = f_prime

Using MATLAB to evaluate the expression, the true value of f'(1) is: 0.3679