Find the derivative of f(x)=sqrt(9−sinx/1−cosx)

Everthing is included in the square root...

To find the derivative of the function f(x) = sqrt(9-sinx/1-cosx), we can use the quotient rule and the chain rule.

1. Rewrite the function:
f(x) = sqrt((9-sinx)/(1-cosx))

2. Apply the quotient rule:
(f/g)' = (f'g - g'f)/(g^2)

Let f(x) = (9-sinx)^(1/2) and g(x) = (1-cosx)^(1/2)

f'(x) = (1/2)*(9-sinx)^(-1/2)*(-cosx)
g'(x) = (1/2)*(1-cosx)^(-1/2)*sinx

3. Apply the chain rule:
(h(g(x)))' = h'(g(x))*g'(x)

Let h(u) = u^(1/2)

h'(u) = (1/2)*u^(-1/2)

4. Combine the results from steps 2 and 3:
f'(x) = [(1/2)*(9-sinx)^(-1/2)*(-cosx)]/[(1-cosx)^(1/2)] - [(1/2)*(1-cosx)^(-1/2)*sinx]*(9-sinx)/(1-cosx)^(3/2)

Simplifying further:
f'(x) = [-cosx*(9-sinx)/(2*(9-sinx)^(1/2)*(1-cosx)^(1/2))] - [sinx*(9-sinx)/(2*(1-cosx)^(3/2))]

This is the derivative of f(x) = sqrt(9-sinx/1-cosx).