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Решите задачу:

$1)\; \; y=ln\sqrt{ \frac{1+sinx}{1-cosx} }\\\\y'= \sqrt{ \frac{1-cosx}{1+sinx} } \cdot \frac{1}{2\sqrt{\frac{1+sinx}{1-cosx}}} \cdot \frac{cosx(1-cosx)-(1+sinx)sinx}{(1-cosx)^2} =\\\\= \frac{1-cosx}{2(1+sinx)} \cdot \frac{cosx-cos^2x-sinx-sin^2x}{(1-cosx)^2} = \frac{cosx-sinx-1}{2(1+sinx)(1-cosx)}\\\\2)\; \; y=log_a(x^2+1)\\\\y'= \frac{2x}{(x^2+1)lna}$

$3)\; \; y=ln(lnx)\\\\y'= \frac{1}{lnx} \cdot \frac{1}{x} \\\\4)\; \; y=ln \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+x} \\\\y'= \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}-x}\cdot \frac{(\frac{2x}{2\sqrt{x^2+1}}-1)(\sqrt{x^2+1}+x)-(\sqrt{x^2+1}-x)(\frac{2x}{2\sqrt{x^2+1}}+1)}{(\sqrt{x^2+1}+x)^2} \\\\5)\; \; y=7^{x^2+2x}\\\\y'=7^{x^2+2x}\cdot ln7\cdot (2x+2)\\\\6)\; \; y=a^{tg\, nx}\\\\y'=a^{tg\, nx}\cdot lna\cdot \frac{n}{cos^2nx}$

$7)\; \; y=(sinx)^{x}\\\\lny=x\cdot ln(sinx)\\\\ \frac{y'}{y}=ln(sinx)+x\cdot \frac{cosx}{sinx} =ln(sinx)+x\cdot ctgx\\\\y'=(sinx)^{x}\cdot (ln(sinx)+x\cdot ctgx)\\\\8)\; \; y= \frac{1-e^{x}}{1+e^{x}} \\\\y'= \frac{-e^{x}(1+e^{x})-(1-e^{x})e^{x}}{(1+e^{x})^2} = \frac{-e^{x}-e^{2x}-e^{x}+e^{2x}}{(1+e^{x})^2} =-\frac{2e^{x}}{(1+e^{x})^2}$