Finding the derivative of y = cos ^ 2x + cos x ^ 2

Finding the derivative of y = cos ^ 2x + cos x ^ 2

y=cos^2x +cosx^2
y'=2cosx(-sinx)+(-sinx^2)*2x
=-2sinxcosx-2xsinx^2
=-sin2x-2xsinx^2

Seek y = (cosx) ^ 2 + cos (x ^ 2) derivative, ask warm-hearted people to help

y'=2cosx*(-sinx)+(-sinx²)*2x
=-sin2x-2xsinx²

Write the derivatives of y = xcosx and y = x part of cosx

y=xcosx
y'=cosx+x(cosx)'
=cosx-xsinx
Y = cosx of X
Y '= x parts (- xsinx cosx)

Given y = ln (2x + 1), we can find the n-th derivative of Y

y'=2/(2x+1)
y''=-4/(2x+1)^2
y'''=16/(2x+1)^3
……
So y (n) = 2 ^ n * (- 1) ^ (n + 1) * (n-2)! / (2x + 1) ^ n