Question

F(x)=x/2x-1; f’(1)-?

Answer 1

Ответ:

Объяснение:

$f(x)=\frac{x}{2x-1}\\f'(x)=\frac{1*(2x-1)-x*2}{(2x-1)^2}=\frac{-1}{(2x-1)^2}\\f(1)=-\frac{1}{1}=-1$

Answer 2

To find $f'(x)$, the derivative of the function $f(x) = \frac{x}{2x-1}$, we can use the quotient rule.

The quotient rule states that if we have a function $f(x) = \frac{g(x)}{h(x)}$, then its derivative $f'(x)$ is given by:

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

In our case, $g(x) = x$ and $h(x) = 2x-1$.

Let's calculate the derivatives:

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

$h'(x) = \frac{d}{dx}(2x-1) = 2$

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

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

Simplify the expression:

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

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

Now, to find $f'(1)$, we simply plug in $x = 1$ into the derived expression:

$f'(1) = \frac{-1}{(2(1) - 1)^2}$

$f'(1) = \frac{-1}{(2 - 1)^2}$

$f'(1) = \frac{-1}{1^2}$

$f'(1) = -1$

So, $f'(1) = -1$.