If x = root sign (1-T Square), y = 1 / Sint, find dy / DX= The answer is dy / DT / DX / dt. DX / DT is the same as the answer, but why does dy / dt calculate 1-T square under 1 / root?

dy/dt=-1/sin ² t · cost
dx/dt=-2t/（2√1-t ²）=- t/√1-t ²
dy/dx=√1-t ² · cost /（t·sin ² t）

Solve the equation: [(square of 1-y under X * root sign)] DX + [square of 1-x under y * root sign] dy = 0

x(1-y^2)^(1/2)dx + y(1-x^2)^(1/2)dy = 0,|x|

Find the special solution of the following differential equation: dy / DX = Y / 2, radical x, y|x = 4 = 1

Find the special dy / DX = Y / (2 √ x) of the following differential equation, y|x = 4 = 1
Separate the variables and get dy / y = DX / (2 √ x);
Take the integral on both sides to get LNY = ∫ DX / (2 √ x) = √ x + C
Substitute the initial condition to get 0 = 2 + C, so C = - 2;
Therefore, the special solution of the original equation is y = e ^ [(√ x) - 2]

Dy / DX is equal to twice the root sign Y, and when x is equal to 0, y is equal to 1, find y (represented by x)

Dy / DX is equal to twice the root sign y
Dy / 2x root y = DX
Integral on both sides:
Root y = x + C
When x equals 0, y equals 1
1=c
Root y = x + 1
y=x^2+2x+1

Given y = ln (1 + x ^ 2), the differential dy at x = 1=

dy=(2x)/(1＋x ²)
Then: dy| (x = 1) = 1

Let XY ln (y + 1) = 0 find the differential dy

xy-ln(y+1)=0
xy=ln(y+1)
Oh, forget it. It's too painful for the computer to play this

If x = root sign (1-T Square), y = 1 / Sint, find dy / DX= The answer is dy / DT / DX / dt. DX / DT is the same as the answer, but why does dy / dt calculate 1-T square under 1 / root?