6.8. $f(x)=3 x^{2}-4 x^{-0.5}+7 \sqrt[3]{x^{2}}-\frac{8}{\sqrt[5]{x^{4}}}$.

6.8. $f(x)=3 x^{2}-4 x^{-0.5}+7 \sqrt[3]{x^{2}}-\frac{8}{\sqrt[5]{x^{4}}}$.

The function is given by:
$f(x)=3 x^{2}-4 x^{-0.5}+7 \sqrt[3]{x^{2}}-\frac{8}{\sqrt[5]{x^{4}}}$.

This function includes:
- A quadratic term: $3 x^{2}$
- An inverse square root term: $-4 x^{-0.5}$
- A cube root term: $7 \sqrt[3]{x^{2}}$
- A fifth root term in the denominator: $-\frac{8}{\sqrt[5]{x^{4}}}$
Solution.

Let's rewrite the function as follows:

$$
\begin{gathered}
f(x)=3 x^{2}-4 x^{-0.5}+7 x^{2 / 3}-8 x^{-4 / 5} \\
f^{\prime}(x)=3 \cdot 2 x^{2-1}-4 \cdot(-0.5) x^{-0.5-1}+7 \cdot \frac{2}{3} x^{\frac{2}{3}-1}-8 \cdot\left(-\frac{4}{5}\right) x^{-\frac{4}{5}-1}= \\
=6 x+\frac{2}{\sqrt{x^{3}}}+\frac{32}{5 \sqrt[5]{x^{9}}}+\frac{14}{3} \cdot \frac{1}{\sqrt[3]{x}}
\end{gathered}
$$

The derivative is defined for $x \in(0 ;+\infty)$.