Question

Differentiate. f(x) = (x3 - 8)2/3 O f'(x) 2x 3 Vx3-8 O f'(x) = S 2x2 f(x) = 3 Vx3-5 O x2 f'(x) 3 Nx3-S

Answer 1

According to the given data we have the following expression:

$f(x)=\mleft(x^3-8\mright)^{\frac{2}{3}}$

To find f'(x) we would have to make the following calculation:

First:

Apply the chain rule

So:

$\mathrm{Apply\: the\: chain\: rule}\colon\frac{d}{dx}\mleft(\mleft(x^3-8\mright)^{\frac{2}{3}}\mright)=\quad \frac{2}{3\left(x^3-8\right)^{\frac{1}{3}}}\frac{d}{dx}\mleft(x^3-8\mright)$$\frac{d}{dx}\mleft(x^3-8\mright)=3x^2$

So:

$\frac{2}{3\left(x^3-8\right)^{\frac{1}{3}}}\frac{d}{dx}\mleft(x^3-8\mright)=\frac{2}{3\left(x^3-8\right)^{\frac{1}{3}}}\cdot\: 3x^2$

Next, you would have to simplify the expression above:

$\mathrm{Simplify\: }\frac{2}{3\left(x^3-8\right)^{\frac{1}{3}}}\cdot\: 3x^2=\quad \frac{2x^2}{\left(x^3-8\right)^{\frac{1}{3}}}$

Therefore:

$\frac{2x^2}{\left(x^3-8\right)^{\frac{1}{3}}}=\frac{2x^2}{\sqrt[3]{x^3}^{}-8}$

Therefore, the right answer is

$f^{\prime}(x)=\frac{2x^2}{\sqrt[3]{x^3}^{}-8}$