Question

производные тригометрические функции​

Answer 1

(sinu)'=u'*cosu

(cosu)'=-u'*sinu

(tgu)'=u'/cos²u

(ctgu)'=-u'/sin²u

(const)'=c'=0

(xⁿ)'=n*xⁿ⁻¹

(u*v)'=u'v+uv'

(c*f(x))'=c*(f(x))'

17.3

a) (f(x))'=2sin2x+2cos(2x);

б)  (f(x))'= 3-4sin(4x)

в)  (f(x))'= 3x²-4cos(2x)

г) (f(x))'= 4/cos²(2x)

17.4

a)  (f(x))'=( 3/sin²x)-12x²

б) (f(x))'=2cos2x+(1/cos²x)

в) (f(x))'= -1/(4cos²x)

г)  (f(x))'=2x*ctgx-(x²/(sin²x))

Answer 2

Ответ:

$17.3)\ \ a)\ \ f(x)=-cos2x+sin2x\\\\f'(x)=2sin2x+2cos2x\\\\b)\ \ f(x)=3x+cos4x\ \ ,\ \ \ f'(x)=3-4sin4x\\\\c)\ \ f(x)=x^3-2sin2x\ \ ,\ \ f'(x)=3x^2-4cos2x\\\\d)\ \ f(x)=2\, tg2x\ \ ,\ \ f'(x)=2\cdot \dfrac{2}{cos^22x}=\dfrac{4}{cos^22x}\\\\\\17.4)\ \ a)\ \ f(x)=-3\, ctgx-4x^3\\\\f'(x)=-3\cdot \dfrac{-1}{sin^2x}-4\cdot 3x^2=\dfrac{3}{sin^2x}-12x^2\\\\b)\ \ f(x)=sin2x+tgx\\\\f'(x)=2\, cos2x+\dfrac{1}{cos^2x}$

$c)\ \ f(x)=4-\dfrac{1}{4}\, tgx\ \ ,\ \ \ f'(x)=0-\dfrac{1}{4}\cdot \dfrac{1}{cos^2x}=-\dfrac{1}{4\, cos^2x}\\\\d)\ \ f(x)=x^2\, ctgx\ \ ,\ \ \ (uv)'=u'v+uv'\\\\f'(x)=2x\cdot ctgx-x^2\cdot \dfrac{1}{sin^2x}$