2.4 Lösning 4b

$$y = {x^2\,\sqrt{x}\over 5} = {1 \over 5}\cdot x^2\,\sqrt{x} = {1 \over 5}\cdot x^2\cdot x^{1 \over 2} = {1 \over 5}\cdot x^{2+{1 \over 2}} = {1 \over 5}\cdot x^{5 \over 2}$$

$$y\,' = {5 \over 2}\cdot {1 \over 5}\cdot x^{{5 \over 2}-1} = {1 \over 2}\cdot x^{3 \over 2} = {1 \over 2}\cdot \sqrt{x^3} = {1 \over 2}\cdot x\,\sqrt{x} = {x\,\sqrt{x}\over 2}$$