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

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Answer 1 · 2016-04-12T12:57:24

$2sinxcosx-6sin2x$

$[1]" "d/dx(3cos2x+sin^2x)$

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

Multiplication by constant: $d/dx[c*f(x)]=c*d/dx[f(x)]$

$[2]" "=3*d/dx(cos2x)+d/dx(sin^2x)$

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

$[3]" "=3*(-sin2x)*d/dx(2x)+d/dx(sin^2x)$

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

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

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

$[5]" "=-6sin2x+2*d/dx(sinx)$

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

$[6]" "=-6sin2x+(2sinx)*(cosx)$

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

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