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)

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

2)

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

3)

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

4)

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

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