Задача 57733 математика...
Условие
Решение
f`(x)=(-x^2+x^3-x^4+x^5)`=-(x^2)`+(x^3)`-(x^4)`+(x^5)`=-2x+3x^2-4x^3+5x^4;
2.
f`(x)=(3x^4-2x^2+4x-1)`=3*(x^4)`-2*(x^2)`+4*(x)`-(1)`=3*4x^3-2*2x+4*1-0=12x^3-4x+4;
3.
f`(x)=(2x^3-4x^7-5x+3)`=2*(x^3)`-4*(x^7)`-5*(x)`+(1)`=2*3x^2-4*7x^6-5*1+0=6x^2-28x^6-5;
4.
y`=([m]\frac{1}{\sqrt[3]{x}})`+\sqrt{x}-7)`=(x^{-\frac{1}{3}}+x^{\frac{1}{2}}-7)`=(x^{-\frac{1}{3}})`+(x^{\frac{1}{2}})`-(7)`=[/m]
[m]=-\frac{1}{3}\cdot x^{-\frac{1}{3}-1}+\frac{1}{2}\cdot x^{\frac{1}{2}-1}+0=[/m]
[m]=-\frac{1}{3}\cdot x^{-\frac{4}{3}}+\frac{1}{2}\cdot x^{-\frac{1}{2}}=[/m]
[m]=-\frac{1}{3\sqrt[3]{x^{4}}}+\frac{1}{2\sqrt{x}};[/m]
5.
(u*v)`=u`*v+u*v`
y`=(ctgx)`*(x^2+6)+ctgx*(x^2+6)`=[m]-\frac{1}{sin^2x}[/m]*(x^2+6)+ctgx*(2x)=[m]-\frac{x^2+6}{sin^2x}[/m]+2x*ctgx;
6.
(u/v)`=(u`*v-u*v`)/(v^2)
y`=((sinx)`*x^3-(sinx)*(x^3)`)/(x^3)^2=((cosx)*x^3-(sinx)*(3x^2))/x^6=(x*cosx-3sinx)/x^([b]4[/b])