Question

Найти производную функции:
$y = ln(cos^{2}x + sqrt{1 + cos^{4}x})$,
вводя промежуточное переменное $u = cos^{2}x$.

Answer 1

$displaystyle y = ln(cos^{2}x+sqrt{1+cos^{4}x})=ln(u+sqrt{1+u^2});\y'=frac{(u+sqrt{1+u^2})'}{u+sqrt{1+u^2}}=frac{u'+frac{(1+u^2)'}{2sqrt{1+u^2}} }{u+sqrt{1+u^2}}=frac{u'+frac{2u'}{2sqrt{1+u^2}} }{u+sqrt{1+u^2}}=\frac{frac{u'(2sqrt{1+u^2})+2u'}{2sqrt{1+u^2}} }{u+sqrt{1+u^2}}=frac{u'(2sqrt{1+u^2})+2u'}{(u+sqrt{1+u^2})2sqrt{1+u^2}}=\frac{u'(2sqrt{1+u^2}+2)}{2usqrt{1+u^2}+2+2u^2}$

$displaystyle frac{(cos^2x)'(sqrt{1+cos^4x}+1)}{cos^2xsqrt{1+cos^4x}+cos^4x+1}\frac{-2cosxsinx(sqrt{1+cos^4x}+1)}{cos^2xsqrt{1+cos^4x}+cos^4x+1}$