设f(t)是二次可微函数且f''(t)不等于0 x=f'(t),y=tf'(t)-f(t),求dy/dx,d^2y/dx^2
人气:403 ℃ 时间:2020-06-12 10:15:20
解答
dx/dt=f''(t)dy/dt=f'(t)+tf''(t)-f'(t)=tf''(t)dy/dx=(dy/dt)/(dx/dt)=1/td^2y/dt^2=f''(t)+tf'''(t)d^2y/dx^2=(d^2y/dt^2)/[(dx/dt)*(dx/dt)]=1/f''(t)+tf'''(t)/[f''(t)^2]
推荐
- f(x)在[0,+∞)内连续,且lim(x→+∞)f(x)=1.证明函数y=e^(-x)∫(0,x)e^tf(t)dt满足方程dy/dx+y=f(x)
- 设函数y=y(x)由x=1-e^t和y=t+e^-t确定,求dy/dx和d^2y/dx^2
- 设f(x)连续,Y=∫0~X tf(x^2-t^2)dt 则dy/dx=?
- 设y=f(x,t),而t是方程F(x,y,t)=0所确定的x,y的函数(F't(x,y,t)≠0),求dy/dx..
- 若函数x=x(t),y=y(t)对t可道6,且x’(t)不等于0,又x=x(t)的反函数存在且可导,则dy除以dx=
- 函数y=log1/2(12-4x-x²)的递增区间是?
- 什么动物的名字是五个字
- 若x分之x的绝对值=1,则x是什么数?
猜你喜欢