Question

найти производные функции срочно пж​

Answer 1

Ответ:

$f(x)=3x^4-7x^3+2x^2+\pi \ \ ,\ \ \ f'(x)=12x^3-21x^2+4x\\\\f(x)=x^5+\sqrt2\ \ ,\ \ \ \ f'(x)=5x^4\\\\f(x)=x^{-6}+2x\ \ ,\ \ \ f'(x)=-6x^{-7}+2\\\\g(x)=2sinx\ \ ,\ \ g'(x)=2cosx\\\\f(x)=sin3x\ \ ,\ \ \ f'(x)=3\, cos3x\\\\f(x)=\dfrac{1}{6}\, cos\, 6x\ \ ,\ \ \ f'(x)=-sin6x\\\\f(x)=5\, tgx\ \ ,\ \ \ f'(x)=\dfrac{5}{cos^2x}\\\\g(x)=tg\, 3x\ \ ,\ \ g'(x)=\dfrac{3}{cos^23x}\\\\f(x)=2(tg5x+10)\ \ ,\ \ \ f'(x)=2\cdot \dfrac{5}{cos^25x}=\dfrac{10}{cos^25x}\\\\g(x)=ctg2x\ \ ,\ \ \ g'(x)=-\dfrac{2}{sin^22x}$

$f(x)=4\, ctg\Big(\dfrac{1}{4}\, x+1\Big)\ \ ,\ \ \ f'(x)=-\dfrac{1}{sin^2\Big(\dfrac{1}{4}\, x+1\Big)}\\\\g(x)=cos(5-3x)\ \ ,\ \ \ g'(x)=3\, sin(5-3x)\\\\f(x)=sin(3-2x)\ \ ,\ \ \ f'(x)=-2\, cos(3-2x)\\\\g(x)=ctg(2-5x)\ \ ,\ \ \ g'(x)=-\dfrac{-5}{sin^2(2-5x)}=\dfrac{5}{sin^2(2-5x)}$