1题:求下列函数在给定点处的导数.
1个回答

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