Find indefinite integral (sin radical X / radical x) DX

Find indefinite integral (sin radical X / radical x) DX

∫ (sin√x)/√x dx
= ∫ 2(sin√x)/(2√x) dx
= 2∫ sin√x d(√x),d(√x) = 1/(2√x) dx
= 2 · (- cos√x) + C
=- 2cos √ x, it's OK to change u = √ x, but this is very simple

Find the derivative dy / DX of the function determined by the parametric equation x=z(1-sinz) y=zcosz

x'z=1-sinz-zcosz
y'z=cosz-zsinz
dy/dz=y'z/x'z=(cosz-zsinz)/(1-sinz-zcosz)

Why does the second derivative of Y versus x = dy / DX versus t ÷ x versus t ÷ why do you want ÷ x versus t

Derivative of dy / DX to t
This is just the derivative of the first derivative with respect to t, not the derivative with respect to x, so it's definitely wrong for light to be equal to this
And multiply by the derivative of t over x, that is
Divide by the derivative of X over t

X = e ^ 2T. Cost y = e ^ t. Sint the value of dy / DX when t = Wu / 6

X = e ^ 2T · cost, y = e ^ t · Sint if that's what your question means, do this:
dx=e^2t·2·cost-e^2t·sint
dy=e^t·sint+ e^t·cost
dy/dx=（e^t·sint+ e^t·cost）/(e^2t·2·cost-e^2t·sint)
It's good to make an appointment and then calculate the value

Let x = Sint, y = cost, then dy / DX=

dy/dt=-sint
dx/dt=cost
∴dy/dx=-sint/cost=-tant

How to explain the problem of derivation of parametric equation? D ^ 2Y / DX ^ 2 = [D / dt (dy / DX)] / DX / DT Why is the second-order reciprocal written as D ^ 2Y / DX ^ 2, why not dy ^ 2 / DX ^ 2, or (dy / DX) ^ 2

First derivative y '= dy / DX
The second derivative Y "= dy '/ DX = D (dy / DX) / DX = D ^ 2Y / DX ^ 2, where there are molecules with two D and one y, so it's a habit to write d ^ 2Y. It's wrong to write (dy / DX) ^ 2, so it's y" = (y') ^ 2
For parametric equations:
x=x(t)
y=y(t)
y'=dy/dx=(dy/dt)/(dx/dt)
The second derivative is also regarded as a parametric equation:
x=x(t)
u=y'=(dy/dt)/(dx/dt)=p(t)
It is also derived from the above parametric equation; y"=du/dx=(dp/dt)/(dx/dt)