Найти производную функции: 1)3^(cos^2(x)); 2)in in in x^2; 3)in((1-sinx)/(1+sinx))^(1/2) !

Ответы

kdfrolova 07.10.2020 08:32

$1)\; \; y=3^{cos^2x}\; ,\; \; (a^{u})'=a^{u}\cdot lna\cdot u'\\\\y'=3^{cos^2x}\cdot ln3\cdot 2cosx\cdot (-sinx)=- ln3\cdot sin2x\cdot 3^{cos^2x}\\\\2)\; \; y=ln\, ln\, ln\, x^2\; ,\; \; y=ln\Big (\underbrace{ln(lnx^2)}_{u}\Big )\; ,\; \; (lnu)'=\frac{1}{u}}\cdot u'\\\\y'= \frac{1}{ln(lnx^2)}\cdot (ln(lnx^2))'= \frac{1}{ln(lnx^2)} \cdot \frac{1}{lnx^2}\cdot (lnx^2)'=\\\\= \frac{1}{ln(lnx^2)}\cdot \frac{1}{lnx^2} \cdot \frac{1}{x^2}\cdot (x^2)'= \frac{1}{ln(lnx^2)}\cdot \frac{1}{lnx^2} \cdot \frac{1}{x^2}\cdot 2x\\\\3)\; \; y=ln \Big (\frac{1-sinx}{1+sinx} \Big )^{1/2}\; ,\; \; y=ln \sqrt{ \frac{1-sinx}{1+sinx}}\; ,\; \; \; (\sqrt{u})'=\frac{1}{2\sqrt{u}}\cdot u'$

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

ПОКАЗАТЬ ОТВЕТЫ

Minikotik2007 07.10.2020 08:32

$\displaystyle 1)\quad y=3^{\cos^2x}\\\\y'=3^{\cos ^2x}*\ln3*(\cos^2x)'=\ln3*3^{\cos^2x}*2\cos x*(\cos x )'=\\\\=-\ln3*3^{\cos^2x}*2 \cos x*\sin x=\boxed{-\ln3*\sin2x*3^{\cos^2x}}\\\\\\\\2)\quad y=\ln \ln \ln x^2\\\\y'=\frac{1}{\ln\ln x^2}*(\ln \ln x^2)'=\frac{1}{\ln \ln x^2}*\frac{1}{\ln x^2}*(\ln x^2)'=\\\\=\frac{1}{\ln \ln x^2}*\frac{1}{\ln x^2}*\frac{1}{x^2}*(x^2)'=\frac{2x}{x^2*\ln x^2* \ln \ln x^2}=\\\\=\boxed{\frac{2}{x\ln x^2* \ln \ln x^2}}\\\\\\\\3)\quad y = \ln\sqrt{\frac{1-sinx}{1+sinx}}$

$\displaystyle y' =\sqrt{\frac{1+sinx}{1-sinx}}*\bigg(\sqrt{\frac{1-sinx}{1+sinx}}\bigg)'=\sqrt{\frac{1+sinx}{1-sinx}}*\frac{1}2*\bigg(\frac{1-sinx}{1+sinx}\bigg)^{-\frac{1}2}*\\\\\\ *\bigg(\frac{1-sinx}{1+sinx}\bigg)'=\frac{1}2*\sqrt{\frac{1+sinx}{1-sinx}}*\sqrt{\frac{1+sinx}{1-sinx}}*\\\\\\ *\bigg(\frac{-cosx(1+sinx)-cosx(1-sinx)}{(1+sinx)^2}\bigg)=\frac{1+sinx}{2(1-sinx)}*\\\\\\ *\bigg(\frac{-cosx-cosx*sinx-cosx+cosx*sinx}{(1+sinx)^2}\bigg)=$
$\displaystyle =\frac{-(1+sinx)*2cosx}{2(1-sinx)(1+sinx)^2}=-\frac{cosx}{(1-sinx)(1+sinx)}=\\\\\\=-\frac{cosx}{1-sin^2x}=-\frac{cosx}{cos^2x}=\boxed{-\frac{1}{cosx}}$