Find the derivative dy / DX of the following function Y = f (sin ^ 2 x) + F (COS ^ 2 x), where f (x) is differentiable

Find the derivative dy / DX of the following function Y = f (sin ^ 2 x) + F (COS ^ 2 x), where f (x) is differentiable

y'=[f(sin^2 x)]'+[f(cos^2 x)]'=f'(sin^2 x)*(sin^2 x)'+f'(cos^2 x)*(cos^2 x)'=f'(sin^2 x)*2sinx*(sinx)'+f'(cos^2 x)*2cosx*(cosx)'=f'(sin^2 x)*2sinx*cosx+f'(cos^2 x)*2cosx*(-sinx)=sin2xf'(sin^2 x)-sin2x...

Solving differential equation (x + y) DX + (3x + 3y-4) dy = 0 Answer x + 3Y + 2ln|x + y-2| = C

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Find the general solution of equation (x + y) DX + (3x + 3y-4) dy = 0

Let u = x + y,
Then DX = Du Dy, substitute into the original equation to obtain:
u(du-dy)+dy(2u-4)=0
udu-udy+2udy-4dy=0
udu+udy-4dy=0
dy=udu/(4-u)=(u-4+4)du/(4-u)=[-1+4/(4-u)]du
So there is y = - u-4ln|4-u| + C
That is, y = - x-y-4ln|4-x-y| + C
The general solution is: x + 2Y + 4|x + y-4| = C

Find the general solution of homogeneous equation (x + y) DX + (3x-3y-4) dy = 0 X + 3Y + 2ln (2-x-y) = C. I don't know how to get the answer in the end 3 (U-1) / (1 + 4u-3u ^ 2) = DX / x, and then I calculate two - 0.5ln absolute values u ^ 2 + 2 / 3u-1 / 3 + radical 2 ln absolute values..... A very complex formula

When solving the equations x + y = 03x-3y-4 = 0, the solution is: x = 2 / 3, y = - 2 / 3. Let x = X-2 / 3Y = y + 2 / 3. Then the original equation is: (x + y) DX + (3x-3y) dy = 0. That is, dy / DX = (1 + Y / x) / (3Y / x-3). Let Y / x = U. then: dy / DX = x * (DU / DX) + U = (1 + U) / (3u-3). That is, X / DX = ((1 + U) / (3u-3)) - U) / Du. The

Let f (x) be differentiable, y = f (x ^ 2), then dy / DX =?

According to the derivation rule of compound function
dy/dx = [f(x^2)]' =f'(x^2) *(x^2)' = 2xf'(x)

Let y = f (X-Y), where f is differentiable and f '≠ 1, then dy / DX =?

y=f(x-y)
dy/dx=f'(x-y)*d(x-y)
=f'(x-y)*(1-dy/dx)
=f'(x-y)-f'(x-y)*dy/dx
[1+f'(x-y)]dy/dx=f'(x-y)
dy/dx=f'(x-y)/[1+f'(x-y)]