Question

Найти первообразную......

Answer 1

Ответ:

$1)\ \ f(x)=2+\dfrac{3}{x^2}-3x^4\ \ ,\ \ \ f'(x)=-\dfrac{6}{x}-12x^3\\\\\\2)\ \ f(x)=3\, sin(x+\dfrac{\pi}{4})\ \ ,\ \ \ f'(x)=3cos(x+\dfrac{\pi}{4})\\\\\\3)\ \ f(x)=\sqrt{3x+2}\ \ ,\ \ \ f'(x)=\dfrac{3}{2\sqrt{3x+2}}\\\\\\4)\ \ f(x)=\dfrac{3}{cos^23x}\ \ ,\ \ \ f'(x)=\dfrac{-3\cdot 2cos3x\cdot (-sin3x)\cdot 3}{cos^43x}=\dfrac{18\, sin3x}{cos^33x}=$

$f'(\dfrac{\pi}{12})=\dfrac{18\cdot sin\dfrac{\pi}{4}}{cos^3\dfrac{\pi}{4}}=\dfrac{18\cdot \sqrt2}{2\cdot \Big(\dfrac{\sqrt2}{2}\Big)^3}=\dfrac{9\cdot \sqrt2\cdot 2^3}{2\sqrt2}=9\cdot 4=36$