Ответ:

-(2+3x⁵)/x³

28x⁶-2sinx

x+5cos5x

6+3/(2(1-3x)^(3/2))

Пошаговое объяснение:

a. 1/x²-x³=d/dx(1/x²)-d/dx(x³)=-2x⁻³-3x²=-2/x³-3x²=-(2+3x⁵)/x³

b. 4x⁷+2cosx=4*(d/dx(x⁷))+2*d/dx(cosx))=4*7x⁶+2*(-d/dx(x)sinx)=28x⁶-2sinx

d/dx(cosx)=(dcosu)/du * du/dx

u=x

d/du(cosu)=-sinu

d/dx(x)=1

c. x²/2+sin5x=1/2*(d/dx(x²))+d/dx(sin(5x))=1/2*(2x)+cos(5x)*(d/dx(5x))=x+5cos5x

d/dx(sinx5x)=(dcosu)/du * du/dx

u=5x

d/du(cosu)=cosu

d. 1/√(1-3x)+6x=d/dx(1/√(1-3x))+6*d/dx(x)=6*d/dx(x)-(d/dx(1-3x)/(2(1-3x)^(3/2))=6*d/dx(x)-d/dx(1)-3*d/dx(x)*1/(2(1-3x)^(3/2))=6*d/dx(x)+3/(2(1-3x)^(3/2))=6+3/(2(1-3x)^(3/2))

d/dx(1/√(1-3x))=d/du * 1/√u * du/dx

u=1-3x

d/du(1/√u)=-1/(2u^(3/2))