Задача 76490 Номер 9.10 Помогитес решением))...
Условие
Решение
y' = 2x - 3sin x - 5^(x)*ln 5 - 1/cos^2 x
b) y = e^(x)*arctg x
y' = e^(x)*arctg x + e^(x)/(1+x^2) = e^(x)*(arctg x + 1/(1 + x^2))
c) [m]y = \frac{\sqrt{x+2}}{x^2-4}[/m]
[m]y'=\frac{\frac{x^2-4}{2\sqrt{x+2}} - \sqrt{x+2}(2x)}{(x^2-4)^2} = \frac{x(x^2-4) - (x+2)}{2x\sqrt{x+2}(x^2-4)^2} = [/m]
[m]= \frac{x(x-2)(x+2) - (x+2)}{2x\sqrt{x+2}(x-2)(x+2)(x^2-4)} = \frac{x(x-2) - 1}{2x\sqrt{x+2}(x-2)(x^2-4)}[/m]
d) [m]y=\sqrt{\log_5(7x^3-3x)} = \frac{1}{2\sqrt{\log_5(7x^3-3x)}} \cdot \frac{1}{(7x^3-3x)\ln(5)} \cdot (21x^2-3) = [/m]
[m]= \frac{21x^2-3}{2\ln(5)\sqrt{\log_5(7x^3-3x)}(7x^3-3x)}[/m]
e) [m]y = (\cos x)^{5x} = e^{\ln(\cos x)^{5x}} = e^{5x \ln(\cos x)}[/m]
[m]y'=e^{5x \ln(\cos x)} \cdot (5\ln(\cos x) +\frac{5x}{\cos x}(-\sin x)) = (\cos x)^{5x} \cdot (5\ln(\cos x) - 5x \cdot tg\ x)[/m]