设u=ln(sinx/y^0.5),其中x=3t^2,y=(1+t^2)^0.5,求du/dt
人气:339 ℃ 时间:2020-05-23 07:16:16
解答
u=lnsin3t^2 - 0.25ln(1+t^2)
du/dt=6tcos3t^2/sin3t^2-0.5t/(1+t^2)=6t cot(3t^2) - 0.5t/(1+t^2)
推荐
- ∫(0→sinx)ln(1+x)dt
- limx→0,[{(sinx)^2,0}ln(1+t)dt]/x^4
- 设f(x)=∫(0--sinx) ln(1+t^2)dt,g(x)=x^3+tan^4 x,则当x--0时,f(x)是g(x)的什么无穷小
- 确定常数a,b,c的值,使lim(x-0) (ax-sinx)/[∫ ln﹙1+t³﹚/t dt]=c
- lim(x→0)[x^4∕∫(0,sinx)ln(1+t)dt]的值
- what are you going to do tomorrow night?(根据实际回答问题)
- I will come to your party.同义句
- 明月松间照,清泉石上流 选自哪,作者是谁
猜你喜欢