x=acos^2 θ, y=sin^2 θ, Find dy / DX

y=1-x/a,dy/dx=-1/a

Find the general solution dy / DX = sin (X-Y)

∵ dy / DX = sin (X-Y) = = > dy = sin (X-Y) DX = = > DX dy = DX sin (X-Y) DX = = > d (X-Y) = (1-sin (X-Y)) DX = = > d (X-Y) / (1-sin (X-Y)) = DX = = > d (X-Y) / (sin ((X-Y) / 2) - cos ((X-Y) / 2)) ^ 2 = DX (application (sin ((X-Y) / 2) ^ 2 + (COS ((X-Y) / 2) ^ 2 = 1) = = > (SEC ((X-Y) / 2)

Y = sin (x + y) find dy / DX There is also an arctan, Y / x = x / y, to find the derivative

y=sin(x+y)
dy=cos(x+y)(dx+dy)
dy=cos(x+y)dx+cos(x+y)dy
dy/dx=cos(x+y)/(1-cos(x+y))

Derivative of LNX / A

It can be seen as taking the derivative of one part a times LNX, and the result is one part a times one part X

Derivative of X + LNX

(x+lnx)' = x' + (lnx)' = 1+ 1/x

What is the derivative of LNX / x?

y'=[(lnx)'*x-lnx*x']/x ²
=(1/x*x-lnx)/x ²
=(1-lnx)/x ²

x=acos^2 θ, y=sin^2 θ, Find dy / DX