Question

Найти производные.....

Answer 1

Пошаговое объяснение:

д)

$\displaystyle y=e^{x-3}ctgx^2$

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

$\displaystyle y'=e^{x-3}(x-3)'ctgx^2+e^{x-3}*(-\frac{1}{sin^2x^2})*(x^2)'=\\\\=e^{x-3}*1*ctgx^2-e^{x-3} *\frac{2x}{sin^2x^2}=e^{x-3}(ctgx^2-\frac{2x}{sin^2x^2})$

e)

$\displaystyle y=ln\frac{arccosx}{\sqrt{x^2+1} }$

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

$\displaystyle y'=\frac{1}{\frac{arccosx}{\sqrt{x^2+1} }} *(\frac{arccosx}{\sqrt{x^2+1} })'=\\\\=\frac{\sqrt{x^2+1} }{arccosx}*\frac{-\frac{1}{\sqrt{1-x^2} }*\sqrt{x^2+1}-arccosx*\frac{1}{2}(x^2+1)^{-\frac{1}{2} }*(x^2+1)' }{(\sqrt{x^2+1})^2 } \\\\=\frac{\sqrt{x^2+1} }{arccosx}*\frac{-\frac{\sqrt{x^2+1} }{\sqrt{1-x^2} }-arccosx*\frac{2x}{2\sqrt{x^2+1} } }{x^2+1 } =\\\\=-\frac{\frac{\sqrt{x^2+1} }{\sqrt{1-x^2} }+\frac{x\;arccosx}{\sqrt{x^2+1} } }{arccosx\sqrt{x^2+1} } =$

$\displaystyle =-\frac{1}{\sqrt{1-x^2}arccosx } -\frac{x}{x^2+1}$