e^y*dy/dx = -cosxdy/dx = -cosx/e^y4.①∫(4->9) √x*(1+√x) dx=∫(4->9) (√x + x) dx= (2/3)x^(3/2) + x²/2 |(4->9)= [(2/3)(9)^(3/2) + 9²/2] - [(2/3)(4)^(3/2) + 4²/2]
2,f′(x)=dy/dx=-1/x²+1/(2√x)dy==[-1/x²+1/(2√x)]dx3,是的,第一你可以做出y=x|x|在[-1,1]的图像,可以发现它是关于原点对称的,在x轴上方的图像与x轴围成的面积与在x轴下方的图像与x轴围成的面积相等.证明的话∫(上值1,下值-1)x|x|dx=∫(上值1,下值0)x|x|dx+∫...
dy/dx = (sint + cost) / (cost - sint) sint = xe^(-t) cost = ye^(-t)dy/dx = (x + y)/(x - y)②根据原参数式可以得到x^2 = 1 + t^2 tany = tx^2 = 1 + (tany)^2 = 1/(cosy)^2(xcosy)^2=1d(xcosy)^2=02(xcosy)(cosydx - xsinydy)=0cosy dx = xsiny dy...
(xcosy)^2=1d(xcosy)^2=02(xcosy)(cosydx - xsinydy)=0cosy dx = xsiny dydy/dx = cosy/(xsiny) = coty/x ③根据原参数式可以得到dx = costdt dy = -2sin2t = -4sintcostdtdy/dx = -4sintd^2y/dx^2 = d(dy/dx)/dx = (-4sint)'/cost = -4cost/cost = -4④根据...
dy/dx = (sint + cost) / (cost - sint) sint = xe^(-t) cost = ye^(-t)dy/dx = (x + y)/(x - y)②根据原参数式可以得到x^2 = 1 + t^2 tany = tx^2 = 1 + (tany)^2 = 1/(cosy)^2(xcosy)^2=1d(xcosy)^2=02(xcosy)(cosydx - xsinydy)=0cosy dx = xsiny ...