Question

1)упростите выражение :(cos(a) - cos(3a)*(sin(a)+sin(3a))/1-cos(4a)
2)найти производную функций а) у=1/4+4х
б)у=(3-х/2)^-9
в)у=(sin(x/2)-cos(x/2))^2

Answer 1

$1)\; \; \frac{(cosa-cos3a)(sinx+sin3x)}{1-cos4a}=\frac{2\, sin2a\cdot sina\, \cdot \, 2\, sin2a\cdot cosa}{2sin^22a}=\frac{2\, sina\cdot cosa}{1}=sin2a\\\\\\2)\; \; y=\frac{1}{4+4x}\; \; ,\; \; y'=-\frac{1\cdot 4}{(4+4x)^2}=-\frac{1}{4\cdot (1+x)^2}\\\\y=(3-\frac{x}{2})^{-9}\; ,\; \; y'=-9\cdot (3-\frac{x}{2})^{-10}\cdot (-\frac{1}{2})=4,5\cdot (3-\frac{x}{2})^{-10}\\\\y=(sin\frac{x}{2}-cos\frac{x}{2})^2\; \; ,\; \; y'=2(sin\frac{x}{2}-cos\frac{x}{2})\cdot (\frac{1}{2}\, cos\frac{x}{2}+\frac{1}{2}\, sin\frac{x}{2})=$

$=sin^2\frac{x}{2}-cos^2\frac{x}{2}=-cosx\\\\\\\star \; \; cos^2\alpha -sin^2\alpha =cos2\alpha \; \; \star$