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