Could someone please help me find the derivatives to these 2 problems?

Question

1. y = x/(1+cos x)

2. y= cos x / (1+sin x)

Answer 1

quotient rule:
d/dx (u/v) = (u'v - v'u) / v^2

y' = [ d/dx (x)(1 + cos(x)) - d/dx (1 + cos(x)) (x) ] / (1 + cos(x))^2

derivative of cos(x) is -sin(x)

y' = [ 1 * (1 + cos(x)) - (-sin(x) * x) ] / [1 + cos(x)]^2

simplify
y' = [1 + cos(x) + x sin(x) ] / [1 + cos(x) ]^2 <== answer


2) y' = [ d/dx cos(x) (1 + sin(x)) - d/dx(1 + sin(x))cos(x)] / [1 + sin(x)]^2

derivative of cos(x) is -sin(x)
derivative of sin(x) is cos(x)

y' = [ -sin(x) (1 + sin(x)) - cos(x) cos(x) ] / [1 + sin(x)]^2

simplify
y' = (-sin(x) - sin^2(x) - cos^2(x) ] / [1 + sin(x)]^2

facotr out the -1 from the second and third term
y' = ( -sin(x) - 1(sin^2(x) + cos^2(x)) ] / [1 + sin(x)]^2

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

so y' = [ -sin(x) - 1] / [ (1 + sin(x) ]^2

fator out -1 again
y' = -1 [ 1 + sin(x) ] / [1 + sin(x) ]^2

simplify
y' = -1 / [ 1 + sin(x) ] <== answer

hope it helps