Answer:
The answer is "0".
Step-by-step explanation:
Please find the correct question in the attached file.
Given value:
[tex]\to f(x) =\frac{x^2}{x^2+5}\\\\[/tex]
Formula:
[tex]\bold{\frac{d}{dx} \frac{u}{v} = \frac{ v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}}[/tex]
[tex]\to f'(x) =\frac{(x^2+5) 2x -2x^3 (2x)}{(x^2+5)^2}\\\\\to f'(0) =\frac{(0^2+5)2(0) -4(0)^4}{(0^2+5)^2}\\\\\to f'(0) =\frac{(0)(5) -2(0)}{(5)^2}\\\\\to f'(0) =\frac{0 - 0}{(5)^2}\\\\\to f'(0) =\frac{0}{25}\\\\\to f'(0) = 0[/tex]
