Differentiate this y = 1 - cos^2 x
Taking differential with respect to x on both sides d/dx y =d/dx(1-cos^2x) dy/dx = d/dx1 -d/dx(cosx)^2 dy/dx =0 - 2cosx *d/dx cosx. dy/dx= -2cosx*(-sinx) d/dx x dy/dx=2sinxcosx *1 dy/dx=sin2x [d/dx (x) =1] by first principle
y = 1 - cos^2 x
y = 1 - (cos x)^2
dy/dx = -2cosx (-sinx)
= 2 sinx cosx
or sin (2x)
To differentiate the given function y = 1 - cos^2(x), we can use the chain rule, which states that the derivative of a composition of functions is equal to the derivative of the outer function multiplied by the derivative of the inner function.
Let's break down the function into two parts:
1. y = u^2, where u = 1 - cos(x)
2. u = 1 - cos(x)
Now, let's find the derivatives of these two parts individually:
1. dy/du = 2u (derivative of u^2 with respect to u)
2. du/dx = sin(x) (derivative of 1 - cos(x) with respect to x)
Finally, we can find the derivative of y = 1 - cos^2(x) by applying the chain rule:
dy/dx = (dy/du) * (du/dx)
= 2u * sin(x)
= 2(1 - cos(x)) * sin(x)
= 2sin(x) - 2sin(x)cos(x)
Therefore, the derivative of y = 1 - cos^2(x) is 2sin(x) - 2sin(x)cos(x).