Продифференцировать функцию

1.у=(arcsin2x)^ln(x+3)

2.у= $\sqrt \frac{6x+5}{6x-5}$ lg(4x+7)

3.y=(x+1)^8(x-3)^2/ $\sqrt{(x+2)^5$

Ответы

drobdrod 10.10.2020 14:49

$1) \ y = (\arcsin 2x)^{\ln(x+3)}\\\ln y = \ln(\arcsin 2x)^{\ln(x+3)}\\\ln y = \ln(x+3) \cdot \ln(\arcsin 2x)\\(\ln y)' = (\ln(x+3) \cdot \ln(\arcsin 2x))'\\\dfrac{y'}{y} = \dfrac{\ln(\arcsin 2x)}{x+3} + \dfrac{2}{\sqrt{1-x^{2}} \arcsin 2x} \\y'=\bigg(\dfrac{\ln(\arcsin 2x)}{x+3} + \dfrac{2}{\sqrt{1-x^{2}} \arcsin 2x} \bigg)\cdot y \\y' = \bigg(\dfrac{\ln(\arcsin 2x)}{x+3} + \dfrac{2}{\sqrt{1-x^{2}} \arcsin 2x} \bigg)(\arcsin 2x)^{\ln(x+3)}$

$2) \ y = \sqrt{\dfrac{6x+5}{6x-5} } \cdot \lg(4x + 7)\\y ' = \dfrac{1}{{2\sqrt{\dfrac{6x+5}{6x-5}}}}} \cdot \dfrac{6(6x-5) - 6(6x+5)}{(6x-5)^{2}}\cdot \lg(4x + 7) + \dfrac{4}{(4x+7)\ln 10} \cdot \sqrt{\dfrac{6x+5}{6x-5} } =\\\\= -\dfrac{30\sqrt{6x-5}\lg(4x + 7)}{\sqrt{6x+5}(6x-5)^{2}} + \dfrac{4}{(4x+7)\ln 10} \cdot \sqrt{\dfrac{6x+5}{6x-5} }$

$3) \ y = (x+1)^{8(x-3)^{\frac{2}{\sqrt{(x+2)^{5}} } }}\\\ln y = \ln(x+1)^{8(x-3)^{\frac{2}{\sqrt{(x+2)^{5}} } }}\\\ln y = 8(x-3)^{\frac{2}{\sqrt{(x+2)^{5}} } } \cdot \ln(x+1)\\\ln (\ln y) = \ln (8(x-3)^{\frac{2}{\sqrt{(x+2)^{5}} } } \cdot \ln(x+1))\\\ln (\ln y) =\ln (8(x-3)^{\frac{2}{\sqrt{(x+2)^{5}} } }) + \ln (\ln(x+1))\\\ln (\ln y) = \dfrac{2}{\sqrt{(x+2)^{5}} } \cdot \ln (8(x-3)) + \ln (\ln(x+1))\\(\ln (\ln y))' = \bigg(\dfrac{2}{\sqrt{(x+2)^{5}} } \cdot \ln (8(x-3)) + \ln (\ln(x+1)) \bigg)'$

$\dfrac{y'}{y\ln y} = -\dfrac{5\ln (8(x-3))}{(x+2)\sqrt{(x+2)^{5}} } + \dfrac{2}{(x-3)\sqrt{(x+2)^{5}}} + \dfrac{1}{(x+1)\ln(x+1)} \\\\y' = \bigg(-\dfrac{5\ln (8(x-3))}{(x+2)\sqrt{(x+2)^{5}} } + \dfrac{2}{(x-3)\sqrt{(x+2)^{5}}} + \dfrac{1}{(x+1)\ln(x+1)} \bigg)\cdot y\ln y$

$y' = \bigg(-\dfrac{5\ln (8(x-3))}{(x+2)\sqrt{(x+2)^{5}} } + \dfrac{2}{(x-3)\sqrt{(x+2)^{5}}} + \dfrac{1}{(x+1)\ln(x+1)} \bigg)\cdot \\\cdot (x+1)^{8(x-3)^{\frac{2}{\sqrt{(x+2)^{5}} } }} \cdot \ln \bigg((x+1)^{8(x-3)^{\frac{2}{\sqrt{(x+2)^{5}} } }}\bigg)$