Question

найти производную функции используя логарифмическое дифференцирование
$y=\frac{x^{2} e^{\sqrt{x}} cosx}{\sqrt{x+1} }$

Answer 1

$y=\frac{x^2e^{\sqrt{x}}\cos{x}}{\sqrt{x+1}}; D(y):x\geq 0\\\\\ln{y}=\ln{\frac{x^2e^{\sqrt{x}}\cos{x}}{\sqrt{x+1}}};\\\\\ln{y}=\ln{{x^2e^{\sqrt{x}}\cos{x}}}}-\ln{{\sqrt{x+1}} } \\\\\ln{y}=\ln{x^2}+\ln{e^{\sqrt{x}}+\ln{\cos{x}}}-\ln{\sqrt{x+1}};\\\\\ln{y}=2\ln|x|+\sqrt{x}+\ln{\cos{x}}-\ln{\sqrt{x+1}};\\\\x>0\Rightarrow|x|=x;$

$\frac{d(\ln{y})}{dx} =\frac{d(2\ln{x}+\sqrt{x}+\ln{\cos{x}}-\ln{\sqrt{x+1}})}{dx};\\\\\frac{\frac{1}{y} dy}{dx}=\frac{2}{x}+\frac{1}{2\sqrt{x}}-\frac{\sin{x}}{\cos{x}}-\frac{1}{2x+2};\\\\y'=y(\frac{2}{x}+\frac{1}{2\sqrt{x}}-\frac{\sin{x}}{\cos{x}}-\frac{1}{2x+2})$

$\\\\\boxed{y'=\frac{x^2e^{\sqrt{x}}\cos{x}}{\sqrt{x+1}} \Bigg(\frac{3x+4+x\sqrt{x}+\sqrt{x}-2x(x+1)\cot{x}}{2x^2+2x}\Bigg)}$