find the derivatives
1. y=sin^3(2x)(cos2x)
There is only one derivative of a function. I assume that you mean
y = sin^3(2x)*cos(2x)
Treat that as the product of two functions u and v, and use the rule
d/dx(u*v) = u dv/dx + v du/dx
u = sin^3(2x)
du/dx = 6 sin^2(2x)*cos(2x)
(I had to use the 'function of a function' rule twice for that one)
v = cos(2x)
dv/dx = -2 sin(2x)
Now put it all together.
To find the derivative of the given function y = sin^3(2x) * cos(2x), we can use the product rule and the chain rule.
Let's break down the function into its separate parts:
u = sin^3(2x)
v = cos(2x)
Now we can find the derivatives of u and v separately.
Derivative of u:
To find the derivative of u, we apply the chain rule:
du/dx = 3sin^2(2x) * d(sin(2x))/dx
= 3sin^2(2x) * 2cos(2x)
Derivative of v:
The derivative of v is straightforward:
dv/dx = -2sin(2x)
Now we can use the product rule to find the derivative of the given function.
dy/dx = u * (dv/dx) + (du/dx) * v
= sin^3(2x) * cos(2x) * (-2sin(2x)) + (3sin^2(2x) * 2cos(2x)) * cos(2x)
= -2sin^4(2x)cos(2x) - 6sin^2(2x)cos^2(2x)
Therefore, the derivative of y = sin^3(2x) * cos(2x) is dy/dx = -2sin^4(2x)cos(2x) - 6sin^2(2x)cos^2(2x).