Question

Помогите, пожалуйста, найти производные.

Answer 1

Производная частного:

$\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}$

1.

$y=\dfrac{(2x^2-1)\sqrt{1+x^2}}{3x^3}$

$y'=\dfrac{((2x^2-1)\sqrt{1+x^2})'\cdot3x^3-(2x^2-1)\sqrt{1+x^2}\cdot(3x^3)'}{(3x^3)^2}=\\\\=\dfrac{((2x^2-1)'\sqrt{1+x^2}+(2x^2-1)(\sqrt{1+x^2})')\cdot3x^3-(2x^2-1)\sqrt{1+x^2}\cdot9x^2}{9x^6}=\\=\dfrac{(4x\sqrt{1+x^2}+(2x^2-1)\cdot\frac{1}{2\sqrt{1+x^2}}\cdot (1+x^2)')\cdot x-(6x^2-3)\sqrt{1+x^2}}{3x^4}=\\=\dfrac{(4x\sqrt{1+x^2}+(2x^2-1)\cdot\frac{1}{2\sqrt{1+x^2}}\cdot 2x)\cdot x-(6x^2-3)\sqrt{1+x^2}}{3x^4}=$

$=\dfrac{(4x+\frac{2x^3-x}{1+x^2})\cdot x-(6x^2-3)}{3x^4}\sqrt{1+x^2}=\\\\=\dfrac{(\frac{4x+4x^3+2x^3-x}{1+x^2})\cdot x-(6x^2-3)}{3x^4}\sqrt{1+x^2}=\\\\=\dfrac{\frac{6x^4+3x^2}{1+x^2}-(6x^2-3)}{3x^4}\sqrt{1+x^2}=\\\\=\dfrac{6x^4+3x^2-(6x^2-3)(1+x^2)}{3x^4(1+x^2)}\sqrt{1+x^2}=\\\\=\dfrac{6x^4+3x^2-6x^2-6x^4+3+3x^2}{3x^4(1+x^2)}\sqrt{1+x^2}=\\\\=\dfrac{3}{3x^4(1+x^2)}\sqrt{1+x^2}=\dfrac{\sqrt{1+x^2}}{x^4(1+x^2)}=\dfrac{1}{x^4\sqrt{1+x^2}}$

2.

$y=\dfrac{x^2}{2\sqrt{1-3x^4}}$

$y'=\dfrac{(x^2)'\cdot2\sqrt{1-3x^4}-x^2\cdot(2\sqrt{1-3x^4})'}{(2\sqrt{1-3x^4})^2}=\\\\=\dfrac{2x\cdot2\sqrt{1-3x^4}-x^2\cdot\dfrac{2}{2\sqrt{1-3x^4}} \cdot(1-3x^4)'}{4(1-3x^4)}=\\\\=\dfrac{4x\sqrt{1-3x^4}-\dfrac{x^2}{\sqrt{1-3x^4}}\cdot(-12x^3)}{4(1-3x^4)}=$

$=\dfrac{4x\sqrt{1-3x^4}+\dfrac{12x^5}{\sqrt{1-3x^4}}}{4(1-3x^4)}=\dfrac{x\sqrt{1-3x^4}+\dfrac{3x^5}{\sqrt{1-3x^4}}}{1-3x^4}=\\\\=\dfrac{x(1-3x^4)+3x^5}{(1-3x^4)\sqrt{1-3x^4}}=\dfrac{x-3x^5+3x^5}{(1-3x^4)\sqrt{1-3x^4}}=\dfrac{x}{(1-3x^4)^{\frac{3}{2}}}$

3.

$y=\dfrac{x^4-8x^2}{2(x^2-4)}$

$y'=\dfrac{(x^4-8x^2)'\cdot2(x^2-4)-(x^4-8x^2)\cdot(2(x^2-4))'}{(2(x^2-4))^2} =\\\\=\dfrac{(4x^3-16x)\cdot2(x^2-4)-(x^4-8x^2)\cdot 2\cdot2x}{4(x^2-4)^2}=\\\\=\dfrac{8(x^3-4x)(x^2-4)-4x(x^4-8x^2)}{4(x^2-4)^2}=\dfrac{2(x^3-4x)(x^2-4)-x(x^4-8x^2)}{(x^2-4)^2}=\\\\=\dfrac{2x^5-8x^3-8x^3+32x-x^5+8x^3}{(x^2-4)^2}=\dfrac{x^5-8x^3+32x}{(x^2-4)^2}$