請教以下微積分解題~麻煩了!

Question

1.Let f(x,y)=e2x+y2 .Find the value of the partial derivative fxyy(0,0)2.Let f(x,y)=tan-1(x/y). Find the value of the parital derivative fy(1,2)3.Let f(x,y,z)=xyz,x>0. Find the value of the partial derivative fz(2,3,0)4.Evaluate the improper integral ∫無窮大3  1/x2-x-25.求極限lim x(ln2)/(1+lnx)          x-->無窮大6.Evaluate the improper integral ∫無限大負無限大 1/4+x2 dx

Answer 1

1.f(x,y)=e2x+y² fx(x,y)=2e2x+y² fxy(x,y)=2(2y)e2x+y² =4ye2x+y² fxyy(x,y)=4y(2y)e2x+y² +4e2x+y² =8y²e2x+y² +4e2x+y² fxyy(0,0)=4e0=42.f(x,y)=tan-1(x/y)fy(x,y)=(1/(1+(x/y)²))(-x/y²)fy(1,2)=1/(1+(1/2)²))(-1/2²)=(4/5)(-1/4)=-1/53.f(x,y,z)=xyz , x>0fz(x,y,z)= (xyz lnx)(y)fz(2,3,0)= (2(3)(0) ln2)(3)=3ln24.∫3∞ 1/(x²-x-2) dx=∫3∞ (1/3)((1/(x-2))-(1/(x+1))) dx=(1/3)[(ln|x-2|)-(ln|x+1|)]3∞=(1/3)[ln|(x-2)/(x+1)|]3∞=(1/3)[ln1-ln(1/4)]=(1/3)(0-ln(1/4))=(1/3)ln45. lim x(ln2)/(1+lnx)x→∞ =lim e^(lnx(ln2)/(1+lnx))x→∞=lim e((ln2)/(1+lnx))(lnx)x→∞=lim e(ln2)(lnx)/(1+lnx)x→∞=lim e(ln2)/((1/(lnx))+1)x→∞=eln2=26.∫-∞+∞ 1/(4+x²) dx=∫-∞0 1/(4+x²) dx+∫0+∞ 1/(4+x²) dx=[(1/2)tan-1(x/2)]-∞0+[(1/2)tan-1(x/2)]0+∞ =(1/2)(0-(-π/2))+(1/2)((π/2)-0)=π/4 +π/4= π/2