Question

Найдите производную сложной функции:
1)$(x^{3} +2)^{2}$
2)sin2x
3)$sqrt{2x-1}$
4)ctg($3x+frac{p}{2}$)

Answer 1

1)

$frac{d}{dx}left(left(x^3+2right)^2right)=\\| f=u^2,::u=left(x^3+2right)\\=frac{d}{du}left(u^2right)frac{d}{dx}left(x^3+2right)=\\=2u^{2-1}cdot frac{d}{dx}left(x^3right)+frac{d}{dx}left(2right)=\\=2ucdot 3x^{3-1}+0=\\= 2ucdot :3x^2=\\| u=left(x^3+2right)\\=2left(x^3+2right)cdot :3x^2=\\=6x^2left(x^3+2right)$

2)

$frac{d}{dx}(sin (2x))=\\| f=sin left(uright),::u=2x\\=frac{d}{du}left(sin left(uright)right)frac{d}{dx}left(2xright)=\\=cos left(uright)cdot :2=\\| u=2x\\= 2cdot cos (2x)$

3)

$frac{d}{dx}(sqrt{2x-1}) = \\|: f=sqrt{u},::u=2x-1\\=frac{d}{du}(sqrt{u})frac{d}{dx}(2x-1)=\\=frac{d}{du}left(u^{frac{1}{2}}right)cdot(frac{d}{dx}left(2xright)-frac{d}{dx}left(1right))=\\=frac{1}{2}u^{frac{1}{2}-1}cdot (2-0)=\\=frac{1}{2}cdot frac{1}{sqrt{u}}cdot 2=\\=frac{1}{2sqrt{u}}cdot2=\\| : u=2x-1\\=frac{1}{2sqrt{2x-1}}cdot 2=\\=frac{1cdot 2}{2sqrt{2x-1}}=\\=frac{1}{sqrt{2x-1}}$

4)

$frac{d}{dx}(cot left(3x+frac{pi }{2})right)=\\| : f=cot (u),::u=(3x+frac{pi }{2})\\=frac{d}{du}(cot (u))frac{d}{dx}(3x+frac{pi }{2})right)=\\=-frac{1}{sin^2(u)} cdot (frac{d}{dx}(3x)+frac{d}{dx}(frac{pi }{2}))=\\=-frac{1}{sin^2(u)} cdot (3+0)=\\=-frac{1}{sin^2(u)}cdot :3=\\| : u=3x+frac{pi }{2}\\=-frac{3}{sin^2(3x+frac{pi }{2})}=$

$| : sin left(3x+frac{pi }{2}right)=cos left(3xright)sin left(frac{pi }{2}right)+cos left(frac{pi }{2}right)sin left(3xright)=cos left(3xright)+0\\=-frac{3}{cos ^2left(3xright)}$