Question

найдите производную функции
1) y=-8x+x^12
2) y=9x^2+5x^4+15
3) y=9x-3√x
4) y=1/x-7x^-4+10
5) y=-2/x^4-3sinx
6)y=tgx+√x
7)y=(x^4+7)(1+x^5)
8)y=5/x^6+ctgx
9)y=√x(3-4x)
10) y=x^8cosx
11) y=sinx/4x^3
заранее спасибо

Answer 1

Ответ:

1) $\boxed{ y' = 12x^{11} - 8 }$

2) $\boxed{y' = 18x + 20x^{3}}$

3) $\boxed{y' = 9 - \dfrac{3}{2\sqrt{x} }}$

4) $\boxed{y' = 28x^{-5}-x^{-2}}$

5) $\boxed{y' = 8x^{-5} - 3 \cos x}$

6) $\boxed{y' = \dfrac{1}{\cos^{2} x} + \dfrac{1}{2\sqrt{x} }}$

7) $\boxed{y' = 4x^{3} + 9x^{8} + 35x^{4}}$

8) $\boxed{ y' = -30x^{-7} - \dfrac{1}{\sin^{2} x} }$

9) $\boxed{ y' = \dfrac{3 - 12x}{2\sqrt{x} } }$

10) $\boxed{y' = 8x^{7}\cos x - x^{8}\sin x}$

11) $\boxed{y' = \dfrac{x\cos x - 3\sin x}{4x^{4}}}$

Объяснение:

1)

$y = -8x + x^{12}$

$y' = (-8x + x^{12})' = -8 + 12x^{12 - 1} = 12x^{11} - 8$

2)

$y = 9x^{2} + 5x^{4} + 15$

$y' = (9x^{2} + 5x^{4} + 15)' = 18x + 20x^{3}$

3)

$y = 9x - 3\sqrt{x}$

$y' = (9x - 3\sqrt{x})' = 9 - \dfrac{3}{2\sqrt{x} }$

4)

$y = \dfrac{1}{x} - 7x^{-4} + 10$

$y' = \bigg(\dfrac{1}{x} - 7x^{-4} + 10 \bigg) = 28x^{-5}-x^{-2}$

5)

$y = \bigg ( -\dfrac{2}{x^{4}} - 3 \sin x \bigg)$

$y' = \bigg ( -\dfrac{2}{x^{4}} - 3 \sin x \bigg)' = (-2x^{-4} - 3 \sin x)' = 8x^{-5} - 3 \cos x$

6)

$y = \rm tg \ x+ \sqrt{x}$

$y' = (\rm tg \ x+ \sqrt{x})' = \dfrac{1}{cos^{2} x} + \dfrac{1}{2\sqrt{x} }$

7)

$y = (x^{4} + 7)(1 + x^{5}) = x^{4} + x^{9} + 7 + 7x^{5}$

$y' = (x^{4} + x^{9} + 7x^{5} + 7)' = 4x^{3} + 9x^{8} + 35x^{4}$

8)

$y = \dfrac{5}{x^{6}} + \rm ctg \ x =5x^{-6} + \rm ctg \ x$

$y' = (5x^{-6} + \rm ctg \ x)' = -30x^{-7} - \dfrac{1}{\sin^{2} x}$

9)

$y = \sqrt{x} (3 - 4x)$

$y' = (\sqrt{x} (3 - 4x))' = (3 - 4x) (\sqrt{x})' + \sqrt{x} (3 - 4x)' = \dfrac{3 - 4x}{2\sqrt{x} } - 4\sqrt{x} = \dfrac{3 - 12x}{2\sqrt{x} }$

10)

$y = x^{8}\cos x$

$y' = (x^{8}\cos x)' = (x^{8})'\cos x + x^{8}(\cos x)' = 8x^{7}\cos x - x^{8}\sin x$

11)

$y = \dfrac{\sin x}{4x^{3} }$

$y' = \bigg( \dfrac{\sin x}{4x^{3} } \bigg)' = \dfrac{4x^{3}(\sin x)' - \sin x(4x^{3})'}{(4x^{3})^{2}} = \dfrac{4x^{3}\cos x - 12x^{2}\sin x}{16x^{6}} =$

$= \dfrac{4x^{2} (x\cos x - 3\sin x)}{16x^{6}} = \dfrac{x\cos x - 3\sin x}{4x^{4}}$