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 · 2016-04-12T12:57:24

$2 sin x cos x - 6 sin 2 x$ $2sinxcosx-6sin2x$

$[1] \frac{d}{d x} (3 cos 2 x + {sin}^{2} x)$ $[1]" "d/dx(3cos2x+sin^2x)$

Sum rule: $\frac{d}{d x} [f (x) + g (x)] = \frac{d}{d x} [f (x)] + \frac{d}{d x} [g (x)]$ $d/dx[f(x)+g(x)]=d/dx[f(x)]+d/dx[g(x)]$

Multiplication by constant: $\frac{d}{d x} [c \cdot f (x)] = c \cdot \frac{d}{d x} [f (x)]$ $d/dx[c*f(x)]=c*d/dx[f(x)]$

$[2] = 3 \cdot \frac{d}{d x} (cos 2 x) + \frac{d}{d x} ({sin}^{2} x)$ $[2]" "=3*d/dx(cos2x)+d/dx(sin^2x)$

The derivative of $cos (x)$ $cos(x)$ is $- sin (x)$ $-sin(x)$ . You can use that here, but you will have to use chain rule.

$[3] = 3 \cdot (- sin 2 x) \cdot \frac{d}{d x} (2 x) + \frac{d}{d x} ({sin}^{2} x)$ $[3]" "=3*(-sin2x)*d/dx(2x)+d/dx(sin^2x)$

The derivative of $2 x$ $2x$ is only $2$ $2$ .

$[4] = 3 \cdot (- sin 2 x) \cdot 2 + \frac{d}{d x} ({sin}^{2} x)$ $[4]" "=3*(-sin2x)*2+d/dx(sin^2x)$

You can use power rule on ${sin}^{2} x$ $sin^2x$ , but you will have to use chain rule as well.

$[5] = - 6 sin 2 x + 2 \cdot \frac{d}{d x} (sin x)$ $[5]" "=-6sin2x+2*d/dx(sinx)$

The derivative of $sin (x)$ $sin(x)$ is #cos(x).

$[6] = - 6 sin 2 x + (2 sin x) \cdot (cos x)$ $[6]" "=-6sin2x+(2sinx)*(cosx)$

$[7] = 2 sin x cos x - 6 sin 2 x$ $[7]" "=color(blue)(2sinxcosx-6sin2x)$

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