The question is an illustration of differential calculus
The value of f'(0) is \(\mathbf{f'(0) = -\frac{4}{3} ( -1)^{-\frac 13}}\)
The function is given as:
\(\mathbf{f(x) = (x^2 - 2x - 1)^{\frac{2}{3}}}\)
Differentiate f(x) using chain rule.
Let
\(\mathbf{u = x^2 - 2x - 1}\)
Differentiate
\(\mathbf{\frac{du}{dx} = 2x - 2}\)
Substitute \(\mathbf{u = x^2 - 2x - 1}\) in \(\mathbf{f(x) = (x^2 - 2x - 1)^{\frac{2}{3}}}\)
\(\mathbf{f(u) = u^\frac23}\)
Differentiate
\(\mathbf{f'(u) = \frac23u^{-\frac13}}\)
So,
\(\mathbf{\frac{dy}{dx} = \frac{du}{dx} \times f'(u)}\)
This gives
\(\mathbf{\frac{dy}{dx} = (2x - 2) \times \frac 23u^{-\frac 13}}\)
Substitute \(\mathbf{u = x^2 - 2x - 1}\)
\(\mathbf{\frac{dy}{dx} = (2x - 2) \times \frac 23(x^2 -2x -1)^{-\frac 13}}\)
\(\mathbf{\frac{dy}{dx} = \frac 23(2x - 2) (x^2 -2x -1)^{-\frac 13}}\)
So, we have:
\(\mathbf{f'(x) = \frac 23(2x - 2) (x^2 -2x -1)^{-\frac 13}}\)
Substitute 0 for x
\(\mathbf{f'(0) = \frac 23(2 \times 0 - 2) (0^2 -2 \times 0 -1)^{-\frac 13}}\)
\(\mathbf{f'(0) = \frac 23( - 2) (0 -0 -1)^{-\frac 13}}\)
\(\mathbf{f'(0) = \frac 23( - 2) ( -1)^{-\frac 13}}\)
\(\mathbf{f'(0) = -\frac{4}{3} ( -1)^{-\frac 13}}\)
Read more about differentiation at:
https://brainly.com/question/24062595
