Question

How do you find the derivative of `[3 cos 2x + sin^2 x]`?

Answer 1

The derivative is `-4 sin(x) cos(x)`.

Explanation:

You need to use the chain rule for differentiation:

if `f(x) = (u @ v)(x) = u(v(x))`, then `f'(x) = u'(v(x)) * v'(x)`.

With this rule in mind, you can compute the derivative as follows:

`f'(x) = 3 * (-sin(2x)) * 2 + 2 * sin(x) * cos(x)`

`color(white)(xxxx) = -6 sin(2x) + 2 sin(x) cos(x)`

... use the transformation `sin(2x) = 2 sin(x) cos(x)` ...

`color(white)(xxxx) = -12 sin(x) cos(x) + 2 sin(x) cos(x)`

`color(white)(xxxx) = -10 sin(x) cos(x)`