微积分题目求解
。∵dy/dx=(xy²-cosxsinx)/(y(1-x²)) ==>y(1-x²)dy=(xy²-cosxsinx)dx ==>y(1-x²)dy-xy²dx+cosxsinxdx=0 ==>(1-x²)d(y²)-y²d(x²)+sin(2x)dx=0 ==>2(1-x²)d(y²)+2y²
微积分题目求解
(1) g(x) = lim<y→+∞>{y/(1+xy)-[1-ysin(πx/y)]/arcsinx}
= lim<y→+∞>{1/(1/y+x)-[1-πxsin(πx/y)/(πx/y)]/arcsinx}
= 1/x-(1-πx)/arcsinx.
(2) lim<x→0+>g(x) = lim<x→0+>(arctanx-x+πx^2)/(xarctanx)
= lim<x→0+>(arctanx-x+πx^2)/x^2 (0/0型)
= lim<x→0+>[1/(1+x^2)-1+2πx]/(2x) (0/0型)
= lim<x→0+>[-x/(1+x^2)^2+π] = π.
大学数学微积分题目，请写出求解详细过程
设大楼AB,楼梯AC,围墙DE,其中B,E,C三点在一条水平线上，AB⊥BC,DE⊥BC,BE=4，DE=8，设CE=x,则CD=√(64+x^2),
DE∥AB,
∴AC/DC=BC/EC,
∴AC=(4+x)√(64+x^2)/x,记为f(x),
f'(x)={x[√（64+x^2）+x(4+x)/√(64+x^2)]-(4+x)√(64+x^2)}/x^2
=(x^3-256)/[x^2*√(64+x^2)],
当x=2^(8/3)时f'(x)=0,f(x)取最小值[4+2^(8/3)]√[2^(2/3)+1]≈17.29（英尺),为所求。
微积分题目，求解。（带过程）
∫∫ (x²+y²)^(1/3) dxdy
=∫∫ r^(2/3)*r drdθ
=∫[0→π/2]dθ∫[0→2] r^(5/3) dr
=(π/2)(3/8)r^(8/3) |[0→2]
=(π/2)(3/8)2^(8/3)
=(3π/4)2^(2/3)