Question

Найти производные. Математика

Answer 1

Пошаговое объяснение:

ж)

$\displaystyle (u+v)'=u'+v'$

$\displaystyle y=arctg\sqrt{x} +\frac{2}{\sqrt[9]{x^8} } =arctgx^{\frac{1}{2} }+2x^{-\frac{8}{9} }$

$\displaystyle y'=\frac{1}{1+(\sqrt{x} )^2}*(x^{\frac{1}{2} } )'+2*(-\frac{8}{9})x^{-\frac{17}{9} } =\\\\=\frac{1}{1+x}*\frac{1}{2}x^{-\frac{1}{2} }-\frac{16}{9x^{\frac{17}{9} }} =\\\\=\frac{1}{2\sqrt{x} (1+x)} -\frac{16}{9x\sqrt[9]{x^8} }$

з)

$\displaystyle y=7^xtg\frac{6^x}{\sqrt{x^3-9} }$

$\displaystyle (\frac{u}{v} )'=\frac{u'v-uv'}{v^2}$

$\displaystyle (uv)'=u'v+uv'$

$\displaystyle y'=7^xln7*tg\frac{6^x}{\sqrt{x^3-9} } +7^x*\frac{1}{cos^2\frac{6^x}{\sqrt{x^3-9} } } *(\frac{6^x}{\sqrt{x^3-9} } )'=\\\\=7^xln7*tg\frac{6^x}{\sqrt{x^3-9} }+\frac{7^x}{cos^2\frac{6^x}{\sqrt{x^3-9} } }*\frac{6^xln6*\sqrt{x^3-9}-6^x*\frac{1}{2}(x^3-9)^{-\frac{1}{2} }*(x^3-9)' }{x^3-9} =\\\\=7^xln7\;tg\frac{6^x}{\sqrt{x^3-9} } +\frac{7^x}{cos^2\frac{6^x}{\sqrt{x^3-9} } } *\frac{6^xln6\sqrt{x^3-9}-\frac{6^x*3x^2}{2\sqrt{x^3-9} } }{x^3-9}$