Find derivative of f(x)=21 (x+ cos x)^2 using chain rule

f'(x) = 21 2(x+cosx)(1-sinx)
f'(x) = 42(x+cosx)(1-sinx)

Chain rule: f(g(x)) = f'(g(x))g'(x)
Steps:
Ignore the 21 Since all constants remain when taking the derivative of constant x another function.
Take the derivative of (x+cosx)^2 as if it were u^2... Let u = (x+cosx), so it would become 2u, substitute back in and you get 2(x+cosx). [This is your f'(g(x)), f(x) = u^2, g(x) = (x+cosx), put g(x) inside f(x) and you get f(g(x)) = (x+cosx)^2]
Next, you have to find g'(x), as we defined before, g(x) = (x+cosx)... so g'(x) = 1-sin(x).

f'(g(x)) = 2(x+cosx) g'(x) = (1-sinx)
f'(g(x))g'(x) = 2(x+cosx)(1-sinx) x 21 <- don't forget the constant at the beginning.

Simplify: 42(x+cosx)(1-sinx)

2*21(x+cosx)*(1-sinx)

The *(1-sinx) is from the chain rule (derivative of inside function).

21 * [2(1 - sinx)(x + cosx)]

→ 42(1 - sinx)(x + cosx)

→ 42(x + cosx - xsinx - sinxcosx)

→ 42x + 42cosx - 42xsinx - 42sinxcosx

f(x) = 21 ( x+ cosx)^2
f '(x) = 21 d/dx( x+cosx)^2
f'(x) = 21 [ 2(x+ cosx) * d/dx ( x+cosx)
f '(x) = 21[ 2 (x+cosx) * ( 1-sinx) ]
f '(x) = 42 (x+cosx) (1- sinx)

f ` (x) = 42 ( x + cos x ) ( 1 - sin x )