Question

F(z)=√z-1/z; f'(2)найти проізводну

Answer 1

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Решение в файле

Answer 2

To find the derivative of the function $F(z) = \frac{\sqrt{z}-1}{z}$ and then evaluate it at $z = 2$, we need to use the quotient rule.

The quotient rule states that for a function $f(z) = \frac{g(z)}{h(z)}$, its derivative $f'(z)$ is given by:

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

Let's find the derivative of $F(z)$:

First, we identify $g(z) = \sqrt{z}-1$ and $h(z) = z$.

Next, we find their derivatives:

$g'(z) = \frac{d}{dz}(\sqrt{z}-1) = \frac{1}{2\sqrt{z}}$

$h'(z) = \frac{d}{dz}(z) = 1$

Now, we can apply the quotient rule:

$F'(z) = \frac{(\frac{1}{2\sqrt{z}}) \cdot z - (\sqrt{z}-1) \cdot 1}{z^2}$

$F'(z) = \frac{\frac{z}{2\sqrt{z}} - \sqrt{z}+1}{z^2}$

Now, to find $F'(2)$, we plug in $z = 2$ into the derivative:

$F'(2) = \frac{\frac{2}{2\sqrt{2}} - \sqrt{2}+1}{2^2}$

Simplifying further:

$F'(2) = \frac{\frac{2}{2\sqrt{2}} - \frac{2\sqrt{2}}{2\sqrt{2}} + \frac{2\sqrt{2}}{2\sqrt{2}}}{4}$

$F'(2) = \frac{\frac{2-2\sqrt{2}+2\sqrt{2}}{2\sqrt{2}}}{4}$