Answer:
f'(x) = -1/(1 - Cos(x))
Step-by-step explanation:
The quotient rule for derivation is:
For f(x) = h(x)/k(x)
[tex]f'(x) = \frac{h'(x)*k(x) - k'(x)*h(x)}{k^2(x)}[/tex]
In this case, the function is:
f(x) = Sin(x)/(1 + Cos(x))
Then we have:
h(x) = Sin(x)
h'(x) = Cos(x)
And for the denominator:
k(x) = 1 - Cos(x)
k'(x) = -( -Sin(x)) = Sin(x)
Replacing these in the rule, we get:
[tex]f'(x) = \frac{Cos(x)*(1 - Cos(x)) - Sin(x)*Sin(x)}{(1 - Cos(x))^2}[/tex]
Now we can simplify that:
[tex]f'(x) = \frac{Cos(x)*(1 - Cos(x)) - Sin(x)*Sin(x)}{(1 - Cos(x))^2} = \frac{Cos(x) - Cos^2(x) - Sin^2(x)}{(1 - Cos(x))^2}[/tex]
And we know that:
cos^2(x) + sin^2(x) = 1
then:
[tex]f'(x) = \frac{Cos(x)- 1}{(1 - Cos(x))^2} = - \frac{(1 - Cos(x))}{(1 - Cos(x))^2} = \frac{-1}{1 - Cos(x)}[/tex]
