Question

Find the value of the derivative (if it exists) at the indicated extremum. (If an answer does not exist, enter DNE.)

Answer 1

Answer:

The answer is "0".

Step-by-step explanation:

Please find the correct question in the attached file.

Given value:

$\to f(x) =\frac{x^2}{x^2+5}$

Formula:

$\bold{\frac{d}{dx} \frac{u}{v} = \frac{ v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}}$

$\to f'(x) =\frac{(x^2+5) 2x -2x^3 (2x)}{(x^2+5)^2}\\\\\to f'(0) =\frac{(0^2+5)2(0) -4(0)^4}{(0^2+5)^2}\\\\\to f'(0) =\frac{(0)(5) -2(0)}{(5)^2}\\\\\to f'(0) =\frac{0 - 0}{(5)^2}\\\\\to f'(0) =\frac{0}{25}\\\\\to f'(0) = 0$