Вопрос:

Найти производную функции $$y=9x^2+5x^4+15-3\sqrt{x}$$ $$y=-\frac{2}{x^4}-3\sin x$$ $$y=(x^4+7)(1+x^5)$$ $$y=\sqrt{x(3-4x)}$$ $$y=\frac{8x^3}{2x-9}$$ $$y=\frac{4\sqrt{x}}{x^3+5}$$ $$y=\frac{\sin x}{4x^3}$$

Ответ:

Найдём производные функций:

  • \( y = 9x^2 + 5x^4 + 15 - 3\sqrt{x} \)
    \( y' = (9x^2)' + (5x^4)' + (15)' - (3\sqrt{x})' \)
    \( y' = 18x + 20x^3 + 0 - 3 \cdot \frac{1}{2\sqrt{x}} \)
    \( y' = 18x + 20x^3 - \frac{3}{2\sqrt{x}} \)
  • \( y = -\frac{2}{x^4} - 3\sin x \)
    \( y = -2x^{-4} - 3\sin x \)
    \( y' = (-2x^{-4})' - (3\sin x)' \)
    \( y' = -2 \cdot (-4)x^{-5} - 3\cos x \)
    \( y' = 8x^{-5} - 3\cos x = \frac{8}{x^5} - 3\cos x \)
  • \( y = (x^4 + 7)(1 + x^5) \)
    \( y' = (x^4 + 7)'(1 + x^5) + (x^4 + 7)(1 + x^5)' \)
    \( y' = (4x^3)(1 + x^5) + (x^4 + 7)(5x^4) \)
    \( y' = 4x^3 + 4x^8 + 5x^8 + 35x^4 \)
    \( y' = 9x^8 + 35x^4 + 4x^3 \)
  • \( y = \sqrt{x(3-4x)} = \sqrt{3x - 4x^2} \)
    \( y' = \frac{1}{2\sqrt{3x - 4x^2}} \cdot (3x - 4x^2)' \)
    \( y' = \frac{1}{2\sqrt{3x - 4x^2}} \cdot (3 - 8x) \)
    \( y' = \frac{3 - 8x}{2\sqrt{3x - 4x^2}} \)
  • \( y = \frac{8x^3}{2x-9} \)
    \( y' = \frac{(8x^3)'(2x-9) - (8x^3)(2x-9)'}{(2x-9)^2} \)
    \( y' = \frac{24x^2(2x-9) - 8x^3(2)}{(2x-9)^2} \)
    \( y' = \frac{48x^3 - 216x^2 - 16x^3}{(2x-9)^2} \)
    \( y' = \frac{32x^3 - 216x^2}{(2x-9)^2} = \frac{8x^2(4x - 27)}{(2x-9)^2} \)
  • \( y = \frac{4\sqrt{x}}{x^3+5} \)
    \( y' = \frac{(4x^{1/2})'(x^3+5) - (4x^{1/2})(x^3+5)'}{(x^3+5)^2} \)
    \( y' = \frac{(4 \cdot \frac{1}{2}x^{-1/2})(x^3+5) - (4x^{1/2})(3x^2)}{(x^3+5)^2} \)
    \( y' = \frac{(2x^{-1/2})(x^3+5) - 12x^{5/2}}{(x^3+5)^2} \)
    \( y' = \frac{\frac{2x^3+10}{\sqrt{x}} - 12x^{5/2}}{(x^3+5)^2} \)
    \( y' = \frac{2x^3+10 - 12x^3}{(x^3+5)^2 \sqrt{x}} = \frac{10 - 10x^3}{(x^3+5)^2 \sqrt{x}} \)
  • \( y = \frac{\sin x}{4x^3} \)
    \( y' = \frac{(\sin x)'(4x^3) - (\sin x)(4x^3)'}{(4x^3)^2} \)
    \( y' = \frac{\cos x \cdot 4x^3 - \sin x \cdot 12x^2}{16x^6} \)
    \( y' = \frac{4x^3\cos x - 12x^2\sin x}{16x^6} = \frac{x^2(4x\cos x - 12\sin x)}{16x^6} \)
    \( y' = \frac{4x\cos x - 12\sin x}{16x^4} = \frac{x\cos x - 3\sin x}{4x^4} \)

Ответ:
1. \( y' = 18x + 20x^3 - \frac{3}{2\sqrt{x}} \)
2. \( y' = \frac{8}{x^5} - 3\cos x \)
3. \( y' = 9x^8 + 35x^4 + 4x^3 \)
4. \( y' = \frac{3 - 8x}{2\sqrt{3x - 4x^2}} \)
5. \( y' = \frac{32x^3 - 216x^2}{(2x-9)^2} \)
6. \( y' = \frac{10 - 10x^3}{(x^3+5)^2 \sqrt{x}} \)
7. \( y' = \(\frac{x\cos x - 3\sin x}{4x^4}\) \>