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

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Answer 1 · 2015-11-27T10:14:50

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

You need to use the chain rule for differentiation:

if $f (x) = (u \circ v) (x) = u (v (x))$ $f(x) = (u @ v)(x) = u(v(x))$ , 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 $sin(2x) = 2 sin(x) cos(x)$ ...

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

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

How do you find the derivative of [3 cos 2x + sin^2 x][3cos2x+sin2x][3 cos 2x + sin^2 x]?