f(x)=1/3x³+2x²-3x+1
f'(x)=x^2+4x-3
f'(0)=-3,f'(1)=2;
f(x)=x²sinx
f'(x)=2xsinx+x^2cosx.
f'(0)=0.
f'(∏/2)=∏.
y=x³-2√x ̄+1/x+1;
y'=3x^2+1/√x ̄-1/(x^2)
y=xsinxInx
y'=sinxInx+xcosxInx+sinx
=sinx(Inx+1)+xcosxInx
y=xsinx/1+cosx
y'=[(sinx+xcosx)(1+cosx)+x(sinx)^2]/(1+cosx)^2
=[sinx+xcosx+sinxcosx+x]/(1+cosx)^2
ρ=√Θ ̄sinΘ
ρ'=-1/(2√Θ ̄)sinΘ+√Θ ̄cosΘ
y=In(2x)·sin3x
y'=1/x*sin3x+3In(2x)·cos3x
y=In(secx+tanx);
y'=[tanxsecx+(secx)^2]/(secx+tanx)]
y=∏/x²+x²Ina
y'=-2∏/x^3+2xIna