127. Find the derivative of the given function and then calculate its particular value at the specified argument value:
1) $F(x)=\frac{(1-\sqrt{\bar{x}})^{2}}{x}, \quad x=0.01$
2) $z=\frac{\cos t}{1-\sin t}, \quad t=\frac{\pi}{6}$
3) $y=\frac{a+b}{3-2 x}+\frac{5 x^{4}-1}{a-b}, x=0$.
Solution. 1) First, we expand the brackets and perform the division, then differentiate:

$$
\begin{gathered}
F(x)=\frac{1-2 V x+x}{x}=\frac{1}{x}-\frac{2}{V x}+1=x^{-1}-2 x^{-\frac{1}{2}}+1 \\
F^{\prime}(x)=-x^{-2}-2\left(-\frac{1}{2}\right) x^{-\frac{3}{2}}=-\frac{1}{x^{2}}+\frac{1}{\sqrt{x^{3}}}
\end{gathered}
$$

Substituting the value \( x=0.01 \), we get

$$
F^{\prime}(0.01)=-\frac{1}{0.01^{2}}+\frac{1}{\sqrt{0.01^{3}}}=-100^{2}+10^{3}=-9000
$$

2) Using formula 4, we find

$$
\begin{gathered}
z^{\prime}=\frac{(\cos t)^{\prime}(1-\sin t)-\cos t(1-\sin t)^{\prime}}{(1-\sin t)^{2}}= \\
=\frac{-\sin t(1-\sin t)-\cos t(-\cos t)}{(1-\sin t)^{2}}=\frac{-\sin t+\sin ^{2} t+\cos ^{2} t}{(1-\sin t)^{2}}=\frac{1}{1-\sin t}
\end{gathered}
$$

Assuming \( t=\frac{\pi}{6} \), we get \( z^{\prime}\left(\frac{\pi}{6}\right)=\frac{1}{1-0.5}=2 \).

3) Applying formulas 46 and 4a, we get

$$
y^{\prime}=-\frac{(a+b)(3-2 x)}{\left.1^{2}-2 x\right)^{2}}+\frac{\left(5 x^{4}-1\right)^{\prime}}{a-b}=\frac{2(a+b)}{(3-2 x)^{2}}+\frac{20 x^{3}}{a-b}
$$

At \( x=0 \), we find \( y^{\prime}(0)=\frac{2}{9}(a+b) \).

Using the differentiation formulas, find the derivatives of the following functions: