Help me find derivative please...?

Question

I am not doing this just for an answer...I want to fully understand the steps please.

g(x)= cos^2 x + sin^2 x +sin x

Answer 1

g(x) = cos^2x + sin^2x + sinx

trig identity

cos^2 x + sin^2x = 1

g(x) = 1 + sinx

g'(x) = d/dx 1 + d/dx sinx

derivative of a constant is 0
derivative of sinx is cosx

so g'(x) = 0 + cosx
g'(x) = cosx

Answer 2

First, I'd simplify g(x). Since cos^2 + sin^2 always equals 1, you can rewrite it as:
g(x) = 1 + sin x

When you take the derivative, the 1 goes away, so....

g'(x) = cos x

Hope that makes sense to you!

Answer 3

well using trigonometric identities you know that

cos^2 (x) + sin^2 (x) = 1

therefore

g(x) = 1 + sin (x)
and hence
g'(x) = cos (x)

Answer 4

since they are added together you can take the derivative seperately for each trig function

assuming you want g'(x) = (cos x)^2 + (sin x)^2 + sin x

two chain rules and a regular trig function

g'(x) = - 2(sin x)(cos x) + 2(cos x)(sin x) + cos x
g'(x) = cos x