[tex]f'(x) = \frac{\sqrt 5}{2 \sqrt x}\\[/tex]
Given:
[tex]f(x) = \sqrt{5x}[/tex]
Note: You can always comment if I have misinterpreted the Given
Please refer to Part a of my Answer from this Question to know more about the needed information: brainly.com/question/24699388
Solving for [tex]\frac{\mathrm{d}}{\mathrm{d}x}(f(x))\\[/tex]:
[tex]\frac{\mathrm{d}}{\mathrm{d}x}(f(x)) = \frac{\mathrm{d}}{\mathrm{d}x}(\sqrt{5x}) \\ f'(x) = \frac{\mathrm{d}}{\mathrm{d}x}(\sqrt 5 \cdot \sqrt x) \\ f'(x) = \sqrt 5 \cdot \frac{\mathrm{d}}{\mathrm{d}x}(\sqrt x) \\ f'(x) = \sqrt 5 \cdot \frac{\mathrm{d}}{\mathrm{d}x}(x^{\frac{1}{2}}) \\ f'(x) = \sqrt 5 \cdot (\frac{1}{2}x^{\frac{1}{2} -1}) \\ f'(x) = \frac{\sqrt 5}{2}(x^{\frac{1}{2} -\frac{2}{2}}) \\ f'(x) = \frac{\sqrt 5}{2}(x^{-\frac{1}{2}}) \\ f'(x) = \frac{\sqrt 5}{2 \sqrt x}[/tex]
