Find the derivative of y with respect to x:

y=(1+cos²x)^6
y'=6(1+cos²x)^5

How do you derive inside the brackets? The answer says -sin2x, but wouldn't it be -2sinx, using the chain rule.

You still have to differentiate the "inside"

d(1+cos^2x) = 2cosx(-sinx)
= -2sinxcox
= -sin 2x, one of the identities

so
y'=6(1+cos²x)^5(-sin2x)
= y'= -6(1+cos²x)^5(sin2x)

Oh, I see it now, thanks!

To find the derivative of y = (1 + cos²x)⁶, we can use the chain rule.

Let's start by identifying the inner function u = 1 + cos²x and the outer function v = u⁶.

To find the derivative of v with respect to u (dv/du), we can simply multiply the power by the old exponent and reduce the exponent by one. In this case, dv/du = 6u⁵.

Next, we need to find the derivative of the inner function u with respect to x, which requires the chain rule.

u = 1 + cos²x
To differentiate cos²x, we can rewrite it as (cosx)² and apply the chain rule. Let's call cosx = w.

u = 1 + w²
du/dx = (du/dw) * (dw/dx)

Now, we can find du/dw by taking the derivative of u with respect to w, which would be 2w. And dw/dx is the derivative of cosx, which is -sinx. Therefore:

du/dx = (du/dw) * (dw/dx)
du/dx = 2w * (-sinx)
du/dx = -2w * sinx

Finally, we can substitute back u = 1 + cos²x and du/dx = -2w * sinx into the derivative of v with respect to u:

dv/dx = (dv/du) * (du/dx)
dv/dx = 6u⁵ * (-2w * sinx)
dv/dx = -12u⁵ * w * sinx

Since u = 1 + cos²x and w = cosx, we can further simplify:

dv/dx = -12(1 + cos²x)⁵ * cosx * sinx

Therefore, the derivative of y = (1 + cos²x)⁶ with respect to x is given by:

y' = dv/dx = -12(1 + cos²x)⁵ * cosx * sinx

So, the answer is -12(1 + cos²x)⁵ * cosx * sinx, not -sin2x.