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)#