203. Find the derivatives of the following functions:
1) $y=x^{x}$;
2) $r=(\cos \alpha)^{\sin 2 \alpha}$;
3) $s=\frac{2 t}{\sqrt{1-t^{2}}}$;
4) $R=(x-1) \sqrt[3]{(x+1)^{2}(x-2)}$.
Solution. Applying logarithmic differentiation, we sequentially find:

1) a) $\ln y = x \ln x$;

b) $\frac{y'}{y} = (x)' \ln x + x (\ln x)' = \ln x + x \cdot \frac{1}{x} = \ln x + 1$;

c) $y' = y(1 + \ln x) = x^x (1 + \ln x)$.

2) a) $\ln r = \sin 2\alpha \ln \cos \alpha$;

b) $\frac{r'}{r} = (\sin 2\alpha)' \ln \cos \alpha + \sin 2\alpha (\ln \cos \alpha)' = 2 \cos 2\alpha \ln \cos \alpha + \sin 2\alpha \left(-\frac{\sin \alpha}{\cos \alpha}\right) = 2 \cos 2\alpha \ln \cos \alpha - 2 \sin^2 \alpha$;

c) $r' = 2 (\cos 2\alpha \ln \cos \alpha - \sin^2 \alpha) (\cos \alpha)^{\sin 2\alpha}$.

3) a) $\ln s = \ln 2 + \ln t - \frac{1}{2} \ln (1 - t^2)$;

b) $\frac{s'}{s} = \frac{1}{t} - \frac{1}{2} \cdot \frac{-2t}{1 - t^2} = \frac{1}{t} + \frac{t}{1 - t^2} = \frac{1}{t(1 - t^2)}$;

c) $s' = \frac{s}{t(1 - t^2)} = \frac{2t}{t(1 - t^2) \sqrt{1 - t^2}} = \frac{2}{\sqrt{(1 - t^2)^3}}$.

4) a) $\ln R = \ln (x - 1) + \frac{2}{3} \ln (x + 1) + \frac{1}{3} \ln (x - 2)$;

b) $\frac{R'}{R} = \frac{1}{x - 1} + \frac{2}{3(x + 1)} + \frac{1}{3(x - 2)} = \frac{2x^2 - 3x - 1}{(x^3 - 1)(x - 2)}$;

c) $R' = \frac{2x^2 - 3x - 1}{(x^2 - 1)(x - 2)} (x - 1) \sqrt[3]{(x + 1)^2 (x - 2)} = \frac{2x^2 - 3x - 1}{\sqrt[3]{(x + 1)(x - 2)^2}}$.

Find the derivatives of the following functions: