2yy'=4 y'=2/y

When t= 1, y=2t=2 y'= 1, that is, when t= 1, the slope is 1.

There is a reason for this: t= 1 is available, and x= 1 y=2 is the passing point (1, 2).

Then the linear equation is: y-2= 1(x- 1) is simplified to y=x+ 1.

2. With x=arcsin(t- 1), we can get t=sin(x+ 1) (1).

Substitute the formula (1) into y 2 = 2t-t 2.

y^2=(cosx)^2

Then dy/dx=|sin(x)|

3. The decreasing interval is that the first derivative is less than 0.

y'=e^x- 1<； 0

Solution: x

That is, the decreasing interval is: [-(infinity), 0]

I know so much, I hope it will help you;