✔ 最佳答案
(1) if y=1/(6t^6+4t+1)=1/u(t) and dt/dx=2xdy/dx=(dy/dt)*(dt/dx)=-[u'(t)/u^2]*2x=-(36t^5+4)*2x/(6t^6+4t+1)^2=-8x(9t^5+1)/(6t^6+4t+1)^2
(2) f(x)=4/(x+x²), g(x)=x² => g'(x)=2xAns: f'(g(x))= d/dx f(g(x))
f[g(x)]=4/(g+g^2)=4/(x^2+x^4)=4/u.....u=x^2+x^4
f'[g(x)]=df/dx=-4u'/(u^2)=-4(2x+4x^3)/(x^2+x^4)^2=-8x(1+2x^2)/x^4*(1+x^2)^2=-8(1+2x^2)/x^3*(1+x^2)^2
df[g(x)]/dx =d[4/(g+g^2)]/dx=-8(1+2x^2)/x^3*(1+x^2)^2
(3) Consider the function y=u/(x+1), where u=3x²-1(3a) y'=dy/dx=[(x+1)u'-u(x+1)']/(x+1)^2=[(x+1)6x-(3x^2-1)]/(x+1)^2=(3x^2+6x+1)/(x+1)^2
(3b) Find an equation for the tangent line to the graph of y(x) at the point where x=1.Ans:y(1)=(3x^2-1)/(x+1)=2/2=1y'(1)=(3+6+1)/4=5/2=(y-1)/(x-1)=> 5x=2y+3
(4) Find the absolute maximum and absolute minimum of the function y(x)=2x^3+3x²-12x-7 on the interval -3<=x<=3Ans:y'(x)=6x^2+6x-12=6(x+2)(x-1) => x=-2, x=1y"(x)=12x+6y"(-2)=-18<0 => maxy"(1)=18>0 => minmin=y(1)=2+3-12-7=-14max=y(-2)=-16+12+24-7=13=相對max絕對max=y(3)=38>13 => absolute max=y(3)=38
