Question

помогите пожалуйста с алгеброй все а.б.в.г​

Answer 1

Ответ:

Объяснение:

a) f(x)=$(x^3-2x^2+3)^1^7$

f'(x) =17*($x^3-2x^2+3)^1^6 *(3x^2-4x)$

b) f(x) =$\sqrt{1-x^4} + \frac{1}{x^2+3}$

f'(x) = -4x³ /(2*$\sqrt{1-x^4}) - 2x/(x^2+3)^2$

c) f(x)=$\sqrt{4x^2+5}$

$f'(x)=8x/(2*\sqrt{4x^2+5})= 4x/ \sqrt{4x^2+5}$

d) f(x)= $(3-x^3)^5 + \sqrt{2x-7}$

$f '(x)=5(3-x^3)^4*(-3x^2) + 0.5*2 /\sqrt{2x-7}$

$=-15x^2(3-3x^3)^4+ 1/\sqrt{2x-7}$

Answer 2

Объяснение:

a)

$f(x)=(x^3-2x^2+3)^{17}\\\\f'(x)=((x^3-2x^2+3)^{17})'=17*(x^3-2x^2+3)^{17-1}*(x^3-2x^2+3)'=\\\\=17*(x^3-2x^2+3)^{16}*(3x^2-4x)=(x^3-2x+3)^{16}*(51x^2-68x).$

б)

$\displaystyle\\f(x)=\sqrt{1-x^4}+\frac{1}{x^2+3} } \\\\f'(x)=(\sqrt{1-x^4}+\frac{1}{x^2+3} })'=((1-x^4)^{\frac{1}{2}} +(x^2+3)^{-1})'=\\\\=\frac{1}{2}*(1-x^4)^{\frac{1}{2}-1}*(1-x^4)'+(-(x^2+3)^{-1-1}) *(x^2+3)'=\\\\=\frac{1}{2}*(1-x^4)^{-\frac{1}{2}}*(-4x^3)-(x^2+3)^{-2}*2x=-\frac{2x^3}{\sqrt{1-x^4} } -\frac{2x}{(x^2+3)^2}.$

в)

$\displaystyle\\f(x)=\sqrt{4x^2+5} \\\\f'(x)=(\sqrt{4x^2+5})'=\frac{1}{2} *((4x^2+5)^\frac{1}{2} )'=(4x^2+5)^{\frac{1}{2}-1} *(4x^2+5)'=\\\\=\frac{1}{2}* (4x^2+5)^{-\frac{1}{2}}*8x=\frac{4x}{\sqrt{4x^2+5} } .$

г)

$\displaystyle\\f(x)=(3-x^3)^5+\sqrt{2x-7} \\\\f'(x)=((3-x^3)^5+\sqrt{2x-7})'=5*(3-x^3)^{5-1}*(3-x^3)'+((2x-7)^\frac{1}{2} )'=\\\\=5*(3-x^3)^4*(-3x^2)+\frac{1}{2} *(2x-7)^{\frac{1}{2}-1}*(2x-7)'\\\\=-15x^2*(3-x^3)^4+ \frac{2}{2*\sqrt{2x-7} } =\frac{1}{\sqrt{2x-7} }-15x^2*(3-x^3)^4.$