Question

Задана функция f(x)=x/x^2+4 найдите f1

Answer 1

F(1) = 1/1² + 4 = 1+4 = 5

Answer 2

To find the derivative of the function $f(x) = \frac{x}{x^2 + 4}$ with respect to $x$, we can use the quotient rule. The quotient rule states that if $f(x) = \frac{g(x)}{h(x)}$, then the derivative $f'(x)$ is given by:

$f'(x) = \frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{[h(x)]^2}$

where $g'(x)$ and $h'(x)$ are the derivatives of $g(x)$ and $h(x)$, respectively.

In our case, $g(x) = x$ and $h(x) = x^2 + 4$. Now, let's find the derivatives of $g(x)$ and $h(x)$:

$g'(x) = \frac{d}{dx}(x) = 1$

$h'(x) = \frac{d}{dx}(x^2 + 4) = 2x$

Now, we can plug these values into the quotient rule formula:

$f'(x) = \frac{1 \cdot (x^2 + 4) - x \cdot 2x}{(x^2 + 4)^2}$

Simplify the expression:

$f'(x) = \frac{x^2 + 4 - 2x^2}{(x^2 + 4)^2}$

$f'(x) = \frac{-x^2 + 4}{(x^2 + 4)^2}$

So, the derivative $f'(x)$ of the function $f(x) = \frac{x}{x^2 + 4}$ is $f'(x) = \frac{-x^2 + 4}{(x^2 + 4)^2}$.